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ARITHMETIC
AND ITS APPLICATIONS J
DBSIGNED AS A
TEXT BOOK FOR COMMON SCHOOLS,
HIGH SCHOOLS, AND ACADEMIES.
ilY
DANA P. COLBURN,
vmnrciPAL of thb bhodb island state normal sohooii^
P&OTIDBNCB.
PHILADELPHIA:
COWPERTHWAIT & CO.
1871.
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CJOt. d( Hon,^ H.
\
Entered, acoordtDg to Aet Af Oongremb in t^ Tesr 1816. by
1>ANA P. COLBURN,
I ffaa Clerked Off 3t of tlM District Ck>iirt of the Piitrict of Rhode
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PREFACE.
TiCE principles involved in Arithmetic are few, the meth«
ods of applying them many. To be a perfect master of the
subject, a person must possess, —
1. A knowledge of the nature and use of numbers, with
the methods of representing and expressing them.
2. A knowledge of the nature and use of the various
numerical operations, with the methods of indicating and of
performing them.
3. Such mental training and cultivation of the reasoning
powers as shall enable him to understand the conditions of
any given problem, and to determine from them what opera-
tions are necessary to its solution.
The first of these includes every thing belonging to Nota-
tion and Numeration.
The second includes the operations of Addition, Subtraction,
Multiplication, and Division ; to which some would add, as a
separate operation, the Comparison of Numbers, as in Frac-
tions and Ratio.
The third requires a power of grasping the various condi-
tions of a problem, of tracing their relations to each other, and
of finding from them what operations must be performed, and
what new relations determined, to obtain the result required.
They are, however, mutually dependent, so that no perscm
(iii)
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IT PRFiTACE.
can master one without learning much of the others. Count
ing is but addition ; and to understand the nature and use of
the number two, we must know that it equals 1 and 1, two
ones, or two times 1 ; that it is 1 more than 1 ; that if 1 be
taken from it 1 will be left, and so on with the other num-
bers — things which require a knowledge of numerical opera-
tions, and also a power of tracing and appreciating relations.
The operations and exercises included in the first two points
are eminently adapted to give quickness of thought and rapid-
ity of mental action. They are to Arithmetic what a knowl-
edge of the nature and power of letters, and of their combina-
tion into words, is to reading. The processes included in them
may be called the mechanical processes of Ai'ithmetic, and
by practice may and should be made so familiar that the mo-
ment a number or a combination is suggested, the mind can
appreciate it, and determine the result
The third requires and imparts a power of investigation,
of tracing out the relations of cause and efiect, and habits of
accuracy both in thought and expression.
To secure these results, it is necessary that the pupil should
be taught in the simplest as well as in the most complicated
problems to reason for himself; to trace fully and clearly the
connection between the conditions of a problem and the steps
taken in its solution ; to state not only what he does, but why
he does it, and indicate the precise character of the result
obtained by each step. Finally, he must learn to gi*asp the
whole mechanical process before performing any part of it, so
that he may know before writing a figure just what additions,
subtractions, multiplications, divisions, and comparisons he has
to make, and be assured that if made correctly they will lead
to the true result. .
Such a course as this is usually taken in works on Oral
Arithmetic. In studying them the scholar is thrown on his
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PREFACE. T
•wn resources : is compelled to learn principles ; to follow out
rigid reasoning processes and connected trains of thought ; to
examine and know for himself the necessity and the reason for
each step taken, and for each operation performed. The result
is, that the studj gives strength, vigor, and healthful disci*
pline to the mind, and becomes an almost invaluable part of
the educating process.
Why should not the same result follow a similar course in
Written Arithmetic? Aside from the writing of numbers,
there is no difference in the principles involved, in the reason-
ing processes demanded, or in the operations required.
In the preparation of this work, the author has kept these
things in view. He has endeavored to present the subject of
Arithmetic as it lies in his o^^m mind, and without any effort
either to follow or t*" deviate from the course pursued by other
writers. He has aimed to arrange the work in such a way as
to lead those who may study it to understand the principles
which lie at the foundation of the science, to learn to reason
upon them, apply them, and to trace out their connections,
relations, and combinations. He has given very full explana-
tions and illustrations, especially of the fundamental opent-
tions ; he has endeavored every where to state principled
rather than rules ; to tlirow the pupil constantly on his own
resources, and force him to investigate and think for himself.
He has omitted some subjects usually found in school arith-
metics, because they do not belong legitimately to the subje^jt
of Arithmetic, because they are of theoretical rather than of
practical importance, or because they require neither spec'^
explanation nor peculiar exercise of the mind.
He has differed from other authors of school arithmetics in
giving algebraic rather than geometrical explanations of the
principles involved in Square and Cube Roots. In this veay
te has been able to give more rigid demonstrations, and mora
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Vi PREFACE.
full explanations, and at the same time (as he conceives) ij
simplify the subject Moreover the processes are in their
nature so essentially algebraic that by the use of squai*es and
cubical blocks we can do nothing more than illustrate some
of theii* applications.
He has given no answers to his problems, because he be-
lieves that to place them within reach of the pupil is always
injurious.
In the first place, such tests are unpractical, for they can
never be resorted to in the problems of real life. What mer*
chant ever thinks of looking in a text book or a key, or of rely-
ing on his neighbor, to learn whether he has added a column
correctly, drawn a correct balance between the debit and
credit sides of an account, or made a mistake in finding the
amount of a bill ?
When a pupil, having left the school room, performs a prob-
lem of real life, how anxious is he to know whether his result
be correct ! Neither text book nor key can aid him now, and
be is forced to rely on himself and his own investigations to
determine the truth or the falsity of his work. J£ he must al-
ways do this in real life, and if his school course is to be a
preparation for the duties of real life, ought he not to do it as
a learner in school ? Is it right to lead him to rely on such
false tests ?
Besides, the labor of proving an operation is usually as val-
uable arithmetical work as was the labor of performing it, and
will oftentimes make a process or solution appear perfectly
simple and clear, when it would otherwise have seemed ob-
scure and complicated.
Again : the science of Mathem^tiqs, of which Arithmetic is
a branch, is an exact science ; it deals in no uncertainties ; its
reasr nings are always accurate, and, if based on true pren^-
jses, nust always lead to true Results In Arithmetic the
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PREFACE. Vii
pupil may always know that a certain step is a true one, and
one which he has a right to take. He may know whether he
has taken it correctly, and thus be certain of the truth of his
first result. He may be as sure of the truth of his second
step and second result, and of his third and his fourth ; and
when he reaches the end, and obtains his final result, he may
be as sure of the truth of that as of any preceding — so sure
that he will be willing to abide by it, and stake his reputation
upon it. (See page 49.)
Why, then, should not the subject be so presented as to
require the pupil to apply such tests, to determine for himself
the truth and accuracy of hi? processes, and thus to form a
habit of patient investigation and just self-reliance? Why
should he not be from the first thrown on his own resources^
and held strictly responsible for the accuracy of his work ?
Would not such a course, if faithfully followed, almost entirely
prevent the formation of those careless habits which bcholars
80 often acquire ?
The articles on business forms and transactions have been
carefully prepared, with a hope of so presenting them as to
give the student true ideas of their use, and of the relations
and obligations of the parties to them.
The materials were drawn from various sources — from
legal works, from intercourse with business men, and from an
article in Mann and Chase's Arithmetic, published originally
in the Common School Journal. To insure accuracy they
were submitted to the inspection of Abraham Payne, Esq., an
eminent lawyer of this city, to whom I am indebted for some
important suggestions.
The work as a whole resembles all other text books (gooJ
or bad) in this — that it requires a good teacher to teach it
well ; as also in this — that it does not contain exactly the
right kind and amount of exercises to meet the wanta of every
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VUl PREFACE.
school or of every class of scholars. The judicious teacliw
will of course extend the exercises which are too meagre,
abridge those which are too full, and omit those which are not
adapted to the wants of his class. We earnestly beg of him,
however, to notice their arrangement, their gradual char-
acter, and their dependence on each other; and not to pass
any till he has convinced himself tlmt they are inappropri-
ate, or that the scholar is master of the operations which they
involve.
To my former teacher, N. Tillinghast, Esq., for many years
principal of the State Normal School at Bridgewater, Massa-
chusetts, I am more deeply indebted than to any other, or all
others, for the ideas embodied in this work. Many of the
processes were learned under his tuition ; and the training
which laid the foundation for whatever real mathematical
knowledge I may possess, was, in a great measure, received
from him. Only those who have been his pupils can appre-
ciate the value of his instructions, and the justice of this
acknowledgment.
The work in its plan and arrangement is entirely my own,
and for its defects I alone must be held responsible. Such as
it is I present it to the public, with a hope that it may be
found useful.
DANA I: COLBDBN.
Pbovidencb, JuLy^ 1855.
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CONTENTS.
SECTION L
FAGS
»efinition8, 1
n NOTAaiON AND NUMERATION
Definitions of Terms, .... 3
Origin of the Decimal System, . • 4
Nattire of the Decimal System, . . 4
Arabic and Roman Methods, . . 5
Figures, 5
Denomination determined by Place, . 6
Names and PL>sition of Decimal Places, 6
Method of reading Numbers, . . 7
Exerci.^es in writing and reading Num-
bers, 7
Number of Decimal Places unlimited, 8
Division into Periods, .... 8
Exercises on Places and Periods, . . 9
Method of reading Numbers, . . 11
All intervening Places to be filled, . 12
Method of writing Numbers, . . 12
Any Combination may be read by itself, 14
Analysis of Numbers, . . . .15
Comparison of Figures in different
Places, 16
Places at Right of Point, . . ,17
Method of reading Decimal Fractions, 18
Method of wntini; Decimal Fractions, 19
Exercises in determining Position of
Point, 20
Effect of changing Place of Point, . SI
Change of Denomination, . . .23
French Method of Numeration, . . S3
English Method of Numeration, . . 23
Comparison of Methods, . 04
Eiorclstfs. ... . . S4
Reman Method, . S5
ni. TABLES.
Introductory, .
United States Money,
Exercises, • • •
English Money, .
Avoirdupois Weight,
Troy Weight,
Apothecaries' Weight, .
Coui|)aris(m of Weights,
Long Measure,
Cloth Measure, •
Square Measure, •
Cubic Measure, •
Angular Measure, .
Dry Measure, •
Liquid Measure, .
Comparison of Measures,
Table of Time, .
Miscellaneous Table,
French Measures and Weights,
tn
27
98
. 28
. 29
, 30
, 30
. 31
. 3^i
32
.33
, 35
, 36
. 39
.40
40
IV. ADDITION.
Definitions, Illustrations, and ExpU
nations, 41
Simple Addition, .... 49
Compound Addition, ... 44
Compound and Simple Addition com
pared, 45
Methods of Proof, . . , .46
Importance of Accuracy and Certainty, 4S
Problems, 5C
Shorter Method of adding Crmipound
Numbers, 54
Common Method, • • • . .55
Examples for Practice, . . . .56
Addition of several Columns, . .58
Leger Columns, 50
19\
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O^i^ TEN IS.
V. SUBTRACT] ON.
Definittriiis and Illiistrations, . . CI
Method when Problciiis requiro no Ue-
d'lctiun, 61
Methods of Proof, . . . . CQ
Problems requiring no Reduction, . 6^
Einiple Subtractiou, ... 64
Compound Subtraction, . . . 66 i
Problems, 67
The changed Minuend not written, . 68
Subtrahend Figure may be increased, ,69
SJiorter Method of subtracting Com-
pound Numbers, ... 70
Problems, ... . . 71
Subtraction from Left to Right, . . 73
Subtraction of several Numbers at
once, 74
VI. MULTIPLICATION.
l)el7nitions and Illustrations, . . 76
Product not affected by Change in Order
of Factors, . . ... 78
Simple Multiplication — when only one
Factor is a large Number, . , 79
Compound Multiplication, . . .80.
Methods of Pr«of, 81
Problems. Multiplier a single Figure, 82
Reduction Descending, . . .85
Multiplication by Factors, . . .88
Both Factors large Numbers, . . 90
Abbreviated Metliod, . . . .93
VIL DIVISION.
Definitions and Illustrations, . . 97
Methods of Proof, . . . .100
Examples. Quotient a single Figure, lOI
Dividend a large Number, . , . 103
Decimal Fractions in the Quotient, . 104
Compound Division, . . . .105
Pniblems for Solution, . . .106
Reduction Ascending, . . .107
Division by Factors, . . , . ? 10
Divisor a large Number, . . 12
Long Division, 16
Abbreviated Process, . . . .117
VIIL CONTRACTIONS AND MIS-
CELLANEOUS PROBLEMS.
To multiply by 5, 12i, 16J, &c., . 119
One Part of Multiplier a Faetor of
another Part, 120
To divide by 99, 999, &c., . . 121
To divide by 5, 16}, 25, See, . . 122
Miscellaneous Probleran, • 1^
Bills of Goods, 12C
Examples for Practice, ~ . 13if
IX. PROPERTIES OF NUMBERS.
DIVISORS, MULTIPLlija, &c
Definitions, ... .133
Demonstration of Principles, 134
Divisibility by 2, 5, 35, &c., 135
*' " 2,25, 1C5, &:a, . 135
" «' 8, 125, 333i, &c., . 136
" * 9, .... 136
" "3, .... 139
" " 11, . . . . 139
Other Tests, 141
Recapitulation, 141
Definitions of Factors, Powers, &,c., . 142
Method of Factoring, . . . .143
Exercises, 145
Common Divisor, Definitions, fcc, . 146
Greatest Common Divisor by Factors, 146
A more brief Method, .... 147
Factoring not necessary, . .148
Method by Addition and Subtractici', 149
Demonstration of Method by Divis^o.', 150
Application of Method by Divisii-n, 150
Common Multiples, Definitions, tt . 151
Least Comnmn Multiple by FactLU 15?V
Abbreviated Method, . . IW
When Factors cannot easily be found. 15(1
X. FRACTIONS.
Introductory, ^ftc
Definitions of Halves, Thirds, &c., . la.
Fractional Parts, . . . VA
Metliod of writing l*ractions, . . I'St
Explanation of Fractions, . . . I'M
Various Kinds of Fractions, . . 161
Operations on Fractions illustrated, . 16i
Reduction to Improper Fractions, . IGl
Reduction to Whole or Mixed Num-
bers, 163
Miitcellaneoiis Operations, . . .164
One Number a Fractional Part of
another, . ... 167
Other Methods of expressing the Value
of a Fraction, . ... 170
To find a Fractional Part of a Number^ ?70
To multiply by a Fraction, . , 172
Practical Problems, . . . .174
Multiplication and Division of the
Numerator, .... 178
Multiplication of the Denominator, 179
Divi:iioD of the Denominator, . . 179
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CONTENTS.
Recapitulation and Inferencei, . . 180
Multiplication and Division of both
Numerator and Denominator by the
same Number, . . . .180
Lowest I'erms, . • • . 181
Cancellation, 183
Compound Fractions, . . . .186
Second Method, . . . .187
Multiplication of Tractions, . . 188
Reduction of a Vulgar to a Decimal
Fraction, 190
Fractional Part of Denominate Num-
bers, 192
To find a Number from its Fractional
Part, 196
Practical Problems, . . * .197
Division by Fractions, . . .200
Process of Division generalized, . 203
Complex Fractions, .... 205
Other Changes in the Terms of a Frats
tion, 207
Common Denominator, . . .208
Addition aud Subtraction of Fractions, 210
XI. APPUCATIONS OF FOREGOING
PRINCIPLES.
. Introductory Note, .... 213
Miscellaneous Problems, . . . 214
Practice, 218
Xn. RATIO AND PROPORTION.
Definitions and Illustrations of Ratio, 222
Reduction of Ratio, . . 223
Definitions and Illustrations of Pro-
portion, 225
To find a missing Terro» . . 226
Relations of Terms, .... 228
Practical Problems, .... 228
Problems in Compound Proporuon, . 230
XUL DUODECIMAL FRACTIONS.
Definitions, .... • 233
Ftoblems, 234
XIV. ALLIGATION.
Definitions and Explanations, . . 237
Problems, 237
XV. INTEREST AND BUSINESS
PROBLEMS.
Introductory, 240
Definitions, 241
Legal Rate. . . . .242
Interest for 2 Months, 200 Months,
&c, at 6 per cent, .... 343
Recapitulation and Inferences, . . 245
Table showing Interest for convenient
Fractional Parts of 200 Months, 20
Montlia, &c, at 6 per cent., . . 246
duestions on the Table, . . . 247
Applications of the Table, . . .247
Interest for various Times, . . 249
Computation of Time, . . . 259
Interest by Days, .... 235
Interest by Dollars for Months, . . 256
Interest by Dollars for Days, . . 257
When to disregard Cents, . . .258
Interest at various Rates obtained
from that at 6 per cent, . . 259
Interest at various Rates obtained
directly, 260
Promissory Notes, . S64
Negotiable Notes, Indorsements, and
Protests, 266
Joint and Several Notes, . 269
Renewal of Note!>, . . . .270
Banks and Banking, .... 272
English Method of computing Interest, 275
Partial Payments, .... 277
Merchants' Method, . . . .281
To find the Time, . . . .283
Equation of Payments, . . .284
To find Date of Equated Time, . . 287
Equation of Accounts, . . . 290
To find the Principal or Interest from
the Amount, 293
Discount and Present Worth, . . 296
Business Method of Discount, . . 296
To find the Rate, . . . . 300
To find the Principal firom the Interest, 301
Compound Interest,
, .
309
Table for Compound Interest, .
303
Commission,
a •
304
Stocks,.
,
.307
Insurance,
,
. 309
Assessment of Taxes,
.
.310
Orders, .
t .
. 311
Bills of Exchange, .
.
.313
Profit and Loss, .
.316
Partnership,
.
. 322
Partnership on Time,
.
. 324
XVI. POWERS AND ROOTS.
Definitions, 327
Denominations of Squares and Rootn, 328
Division into Periods, . . . .329
Method of forming a Square, 329
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CONTENTS.
Method of extracting the Square Root, 331
Rule for Square Root, with Problems, 332
Square Rout of Fractions, . • .333
Square Root of Decimal Fractions, • 334
Cube Root. — Relation of Cube U>
Root, . . . . . . .335
Division into Periods* .... 335
Method of forming a Cube, • 336
To extract the Cube Root, . . 337
Rule for the Cube Root, . . .339
Cube Root of Ftactions, . « .340
Cube Root of Decimal Fractions, . 340
Rule for extracting a Root of any De-
gree, 341
XVn. MENSURATION.
Polygons, , .... 342
Problems, 344
properties of the Right-angled TiiftA-
fiv, with Problems 946
Solids, . . . . . . 34C
Problems, 348
XVIIL PROGRESSIONS.
. 349
.350
. 350
.351
.351
Arithmetical Progression, .
Problems, .
To find the Sum of a Series,
Problems, .
Geometrical Progression, •
Problems, 353
To find the Sum of a Geometrical
Series, 3S9
Problems, 353
Infinite Decreasing Series, . . 354
XIX. CIRCULATING DECIMALS, 354
XX. MISCELLANEOUS EXAM-
PLBS, . . M
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AEITHMETIC.
SECTION I.
1. Definitions.
1. Quantity is a term applied to whatever may be in*
creased, diminished, or measured.
2. A CONCRETE UNIT is any quantity which may be con«
sidered by itself, and made the measure of other similar,
quantities.
(a.) It may be a single thing, as an apple^ a book^ a pound, a Jbot, a
dollar ; or it may be a collection of single things, as a dozen of apples,
a score of sheep.
3. An ABSTRACT UNIT is the idea or conception of one. or
of unity, without reference to any particular thing or quan-
tity. Any number of abstract units, considered as forming a
single collection or whole, may also be regarded as an ab-
stract unit ; as, one ten, one hundred, one thousand, &c.
(a.) The term unit, when nsed without any limitation, always referg
to a single thing or quantity, or to the abstraction, one.
4. A CONCRETE NUMBER oxprcsscs a single thing or
quantity, considered as a unit of measure ; or it expresses how'
many such units there are in any given quantity; as, om#
yard, three shillings, eight barrels, ten birds.
5. An ABSTRACT NUMBER expresses how many times a'
unit is taken or repeated, without any reference to the nature
of the unit; as, one, eight, twelve.
6. The unit which any number expresses is called the
DNIT OF MEASURE, or the UNIT OP COMPARISON. :
(a.) In an abstract number it is the abstract unit, and in a ooncreftt
1 w
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2 DEFINITIONS.
Bumber it is the unit which the nnmber measures Thus, in $even doiUtn
it is a db/Zor, in twdvt busfids it is a bushel, in Jhrtif-.ifee dozen o/eggt it if
a dozen of eggs, in six tens it is one ten, &c
7. Arithmetic is the sciekce or kumbvbs and art of
KUHERIGAL COMPUTATION.
(a.) It treats of numbers with reference to their nature and use, thcii
properties and relations ; explains the various methods of representing
them ; and includes the theory of all numerical operations, as well ai
the practical methods of performing them.
8. The operatioDs of which numbers are susceptible are
four in number; viz.^ Addition^ Subtraction^ MuUipUcation^
and Division,
9. Among the characters used to indicate numerical opera*
tions or relations are the following : —
(€u) = The sign of equality^ called equaly or equal to, sig-
nifies that the quantities between which it is placed are equal
to each other.
(J>,) + The sign of addition, called plus or and, signifies
that the quantities between which it is placed are to be added
together.
Thus, 7 +4 sa 11, is read, aeeen plus four equal deotn ; or, seven ani^
fiur equal eleven ; and means that seren added to four equal eleven.
(c.) — The sign of subtraction, called minus or less^ signi-
fies that the number following it is to be subtracted.
Illustration, 9 — 6 sa 3, is read, nine minus six, or nine less six, equal
three, and means that nine diminished by six equal three.
(d.) X The sign of mvltiplicatum, called times or muUi"
pUed by, signifies that the numbers between which it is placed
are to be multiplied together.
Illustration, 7 X 5»S5, is read, seventimes Jive equal thirty-five; or,
^mMpUed by five equal thirty-five. ^
(e.) -^ The sign of division, called divided by, means that
the number before it is to be divided by that following it
Illustration. 12 -I- S &= 4, is read, twelve divided by three equal fimr^
and means that twelve divided by three gives four for a quotient
(/•) Division maj also be expressed bj writing the nuBi*
L
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NOIATIGN AND NU)fEHATION. •
bet to be divided above the number by ynhich it S3 (c bt
divided, with a line between them.
12
lUustration. y3B:4, m^ans the tame as 12«*-3«b4; Le^tiiat the
qnutient of twelre divided by three equals four.
(</.) Such expressions as jy^, J^ ^, are usually called fradUcm^ aad
read thus : t¥>€lve thirds^ sixteen fourths^ eeven eighths ; though they may,
with equal correctness, be read as twdve divided by three, sixteen divided
etyfour^ setfen divided by eight. When read as fractions, the number above
the line is called the numerator^ and the number below it the denominetor.
. Sec Section X.
(h,) The more common use of fractions is to ea^press the valve of sunt «r
more such parts as are obtained by dividing a unit into a given number 4ff
equal parts, or, which is the same thing, to express the value of one or mm^
equal parts of such kind that a given number of them will equal a unit.
Thus, ^ (read three fourths) is used to express the value of three sueii
parts as would be obtained by dividing a unit into four equal parts ;
er, in other words, the value of three equal parts of such kind that it
would take four of them to equal a unit
(t.) The numerical value of a fraction is the same, whether we oon«
eider that it expresses a division to be performed, or a certain number
of equal parts, and, in eitlier case, it is obvious that a fraction mii«t equal
unity whenever its numerator equals its denominator. Thus, 1 as a »«
SECTION II.
NOTATION AND NUMERATION.
3. Definition of Terms,
Notation and Numeration treat of the various methodf
if representing and expressing numbers.
(a.) The distinction usually made between notation and num^ratioa
is, that the former treats of the methods of representinn: numbers by
written characters, while the latter treats of the methods of reading
IImb, ot of expressing them ii words.
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'4 w en ATios AXE» jnnfXKATios.
9* MMumU of teprmrtdA^ Xitmhen, cad Ort^ ^ dk
Deeimal SysUmt,
JTorabcrs maj be represented bj Materia l objects or VKbis
sarkj, bj word% and bj figures.
ftf.) As oar ideas of vamhtr m deriTcd pciannlj firom mmH liil o^
|cui. 9» the Mo^ tt;inirsl 2nd ob7u>as nKsfud of eoiBaMEciiescza; tibem
t» odicfs ii fc^ exfaiwring af msaj nek objecti as tbezc are aahs m tka
WUBocr coQSbieruL
f 'I,) B » probable diai ia the caifier Mages af sooetj aaidbcfs vcia
leyiejeat ed c«l j ia this waj. the finders bciag. as a gcaefal thia^. mmkb
vascfaseoanscfs, Tbas, three fiagcis voali be shova as a sjmbol ftr
the mmmhcr three, ire fint^ers fior the aamhfr fire, aad dbe fisgcn «f
kodb heads for the aaariier tea.
(&) Sacfa a aMtbod wocdd aatarallj lead a people anag it to repTO>
iort huge aamberf bf exhibicm^ Ae fiagen of bodb haads as aaaj
tfBes as there are tcss la die aaiaben coandcred, aad or was kana^
dbeai to redum bf teas, woaM \xr the fooiHlasioii for a srsteai of aaa-
hen similar to the one in general use, viuch is knowa as the dscimai.*
4« Nature of the Decimal SytUm of Nwmbert.
{a,) The fondamental idea of the Decimal System is, diat ten sngle
things maj be regardtd as forming a nngle collection or graop ; ten of
dbese groops as ibnning a larger group ; and so on, ten gionps of one
size formioganeirgroopof a larger size, each capable of being regarded
tod dealt with as a single thing or nnit. This idea renders it easj to
fepresent the largest nnmben, bj haring names for each of the first ten
nombefs, and lor each groop formed bj combining ten of the smaller
{b,) In conformitj with it, we might connt thns : ome, teo, tkretj Jtmr^
^ffCf $iXf tevetif dgld, mne^ ten, one and ten^ two and ten^ three and ten, four
and ten, five and ten^ tix and ten, tecen and ten, eight and ten, nine and ten,
two tentf two tens and one, two tens and two, &c^ to nine tens and eigit,
fdne tens and nine, ten tens or one hundred, one hundred and one^ &c^ to nine
hundreds nine tens and eight, nine hundreds nine tens and nine, ten hundreds
9r one thousand, kt,
{e,) Adopting this method, and forming componad words hj drop
ping the conjunction, we should connt from ten thns : one-ten, two-ten^
three'tenffimr4en,five4en, six-ten, seven-ten, eight-ten, nine-ten, two-tens, Iwa*
tens- one, two4ens two, tie,
"* The word dseimal is derired from the Latin word decern, which signh
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ITOTATION AND NUUERATIOir. 9
1<L) Changing the word ten to (een, and dropping the hyphen in
counting from ten to two-tens, we should have oneteen, tuxUeen, threettm^
JburteeUfJioeteenf sixteen^ a lentooi, eighteen^ and nineteen,
(e.) By now changing Jive to Ji/^ Jiree to fAtr, and substituting for
me-teen and tvoo-teen the words deven and twdve^ signifying Respectively
one left and tvso left, (i. e., ten and one left, ten and two left,) we should
Aave the familiar names, eleven, twelve, thirteen, fourteen, Jijleen, gixieen^
teventeen^ eighteen, and nineteen,
(f.) Substituting the syllable ty for tens in the words two-tent^ Mnse-
fe»w, &c^ would give the words twoty, threety, fourty, fivety, iixty, eeventg,
nyhty, and ninety f and again changing the titree to thir^four to fir, and the
fae to Jif, and substituting twen, derived from twain, for two, would give
the familiar names twenty, thirty, fatiy, fifty, sixty, seventy, eighty, and
ninety.
{g,) These changes would enable us to count from two-tens or twenty
thus: tioenty-one, twenty-two, &c., to twenty-nine, thirty, thirty-one, &c^ to
9ne hundred, one hundred and one, &c,
(A.) This method of expressing numbers is the one now in general mo
S, Arabic and Roman Methods of Notation.
(a.) Numbers are usually represented to the eye by charac-
ters called figures, though sometimes by letters of the alphabet.
(&.) Tbd method by figures is called the Arabic Method^
because it was introduced into Europe by the Arabs.
Note. — The Arabs probably obtained it from the Persians, who had
obtained it from the Hindoos. Its origin has never been satisfactorily
determined.
(c.) The method by letters is called the Roman Method^
Y M»use it was used by the ancient Romans.
6. The Figures.
* The figures are ten in number, viz. : —
1 or /^ called one,
2 or S, called two.
8 or 3^ called three.
4 or ^^ called four.
5or^, called /m.
4*
6 or O, called six.
7 or y, called seven.
8 or Oj called eight.
9 or ^^ called nine.
or 0, called tero^ cipher^
nothing, or nou^
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6 NOTATrON AND NUMERATION.
T. 21ie Place of a Figure determines its Denominatifrt*,
Each of these figures represents as many units as its nam«
indicates; but the size or denomination of those units is deter-
mined hy the place or position of the figure with reference to
a period or dot, called the decimal point.
8. Names and Position of the Decimal Mamies.
(a.) The figure immediately at the left of the point repre-
sents ones, or simple units ; the second figure at the left repre-
sents tens, (i. e., units of the denomination or value of ten
ones, or ten simple units ;) the third figure represents hun-
dreds ; the fourth represents thousands, and so on ; the figure
in any place always representing ten times the value it would
represent if it stood one place faither towards the right.
(b.) Hence each place has its peculiar name, the first,
second, third, and fourth places being called, respectively, the
units* place, the tens' place, the hundreds' place, and the thou-
sands' place. Moreover, the position of these places is marked
by the figures occupying them. Hence each figure performs
a double ofiice, viz., it marks a place, and indicates as many
of the denomination of that place as its name indicates.
(c.) The following will illustrate this : —
lllll
0000 •
(«f.) In the above example each zero marks a place, and
shows that there are none of the denomination of the place it
occupies expressed in that place.
(e.) As another illustration, take the expression 2503.
Here e^ch figure marks a place, and denotes as many of the
denomination of that place as its name implies; i. e., the 3
marks the units' place, and shows that there are 3 units ; the
marks the tens' place, and shows that there are no tens ; the
5 marks the hundreds' place, and shows that there £^re 5 hun-
dreds ; and the 2 marks the thousands' place, and shows thc^t
there are 2 thousands. The number is read two thousand fiv4
kundred and three.
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NOTATION AND NUMERATION. 7
NoTS. — The zero is often called an insignificant fiffnro* «">d the other
nine digits significant figures ; bat there is no foundation for the distinc
tion. The zero performs an office precisely similar to that of any other
Qgnre, as the above explanation of the use of the figures used in writing
2503 clearly shows. Even when standing by itself it is as expresslre as
any other figure.
(/.) The decimal point is often omitted in writing numbers ;
but in all such cases it is U7ider stood to belong at the right of
the given number, thus making the right hand figure represent
units.
9. Method of reading Numbers,
In reading numbers expressed by figures we begin at the
left hand, i. e., with the highest denomination.
(a.) 546 = five hundreds, four tens, and six units, and is read Jive
kandred and fmiy-six.
(b.) 398 = three hundreds, nine tens, and eight units, and is read three
hundred and ninety-eight.
(c.) 407 = four hundreds, no tens, and seven units, and is read four
hundred and seven,
(d.) 180= one hundred, eight tens, and no units, and is read one
hundred and eighty,
{e.) 64 or 064 = six tens and four units, and is read sixty-four,
10. Exercises in reading and writing Numbers.
Read the following numbers, and also give the value of
each in hundreds, tens, and units : —
1. 507.
4. 379.
7. 031.
2. 409.
5. 211.
8. 37.
3. 528.
6. 403.
9. 200.
10. Explain the use of each figure used in the above num-
bers, as in the following model : —
Model. — In the first number, 507, the 7 marks the units' place, and
shows that there are 7 units ; the murks the tens' place, and sliows tliat
there are tens ; the 5 marks the hundreds' place, and shows that there
are 5 hundreds.
11. How will you write four hundred and seven?
Ana, By writing 4 in the hundreds' place, in the tens', and 7 in th«
■nits' ; thus, 407.
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• NOTATION AND NUMERATION.
12, How will you write two hundred and seventeen?
13 How will you write eight hundred and forty-one ?
14. How will you write eight hundred and twelve ?
15. How will you write seven hundred and forty-six?
16. How will you write six hundred and ninety-four?
17. How will you write nine hundred and sixty-four?
18. How will you write four hundred and sixty-nine ?
19. How will you write nine hundred?
20. How will you write seven hundred and eighty ?
11 • I7und>er of DecimcU Places wilimited.
Extending these principles, we can take as many places af
we please, 5y giving to the figure in ecuih ten tifne$ the value
it would have if written one place farther to the right. The
names of the places as far as the twenty-fourth are given ia
the following example.
I' 1^ |i i i I L
111 Ili III III III 111 III 1..^
Ill III III 1^1 111 111 III I III -
000,0 0,0 0,0 0,0 0, 0,0 oo.,ooo.
.. Ill III III III III III III III ■ ■
Ua iU iU Hi Ui Hi iU lli;
13. Division into Periods.
(a.) By inspecting the above example it will be seen that the
first three places are occupied by units, tens, and hundreds, —
the second three by thousands, tens of thousands, and hundreds
of thousands, — the third three by millions, tens of millions,
and hundreds of millions, and so on. If the first three places
were called, as they might be with perfect propriety, units,
tens of units, and hundreds of units, and we should divide the
number into periods of three figures each by commas, the first
period would be the period of units, the second the period of
thousands, the third of millic^T^.s, the fourth of billions, ^cc
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NOTATION AND NUMERATION.
9
(5.) The right hand figure in each period expresses unitg
or ones of the denomination of that period, while the second
figure expresses tens, and the third, or left hand figure, ex-
presses hundreds of that denomination.
(c.) This is exhibited in the following table : —
i. L u i
i
V
5
i
i
i I
0*000,0 0, 00 0.000, 00, 000,000.
i
S.i?
II:
6 S fe
I
^i
111 II
lit
I.
i i?i i?f 1^1 I I i
■3. 5i-5 esI IS5i ij-s'
Hi
So
III l|i
g e S
tl's |l-a ill 4S-
ill 111 III l?s
Je £ A
Sil ffl
o o 5
111 Il| III
SSS sSo ggs
g ^ s
13. Exercises to secure Familiarity with the Places and
Periods.
1. What is the name of the first period at the left of the
point? of the second period? of the third? of the fourth? of
the fifth ? of the sixth ? of the seventh ? of the eighth ?
2. What is the name of the period occupying the first, sec-
ond, and third places at the left of the point ?
8. Occupjnng the fourth, fifth, and sixth places ?
4. Occupying the seventh, eighth, and ninth ?
5. Occupying the tenth, eleventh, and twelfth ?
6. Occupying the thirteenth, fourteenth, and fifteenth ?
7. Occupying the sixteenth, seventeenth, and eighteenth ?
8. Occup3ring the nineteenth, twentieth, and twenty-first ?
9. Occupying the twenty-second, twenty-third, and twenty
fourth ?
10. What is the number of the millions* period ?
An», The third period a the left of the point.
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10
NOTATION AND NUMERATION.
11 What is the number of tlie thousands' period ?
12. What is the number of the quintillions' period?
13. What is the number of the trillions' period ?
1 4. What is the number of the units' period ?
15. What is the number of the sextillions' period?
IG. What id the number of the billions' period ?
17. What is the number of the quadrillions' period ?
How many places are there between the point and the
Thousands' period?
Billions* period?
Sextillions' period ?
Trillions' period ?
18. Millions' period ? ] 22.
19. Quintillions' period? 23.
20. Units' period? 24.
21. Quadrillions' period ? 25.
26. In which place of what period would the fourth figure
at the leH of the point be ?
Ans. In the first place of the second or thousands* period.
27. In which place of what period would the seventh fig*
ure at the left of the point be ?
28. Would the tenth ?
29. Would the sixteenth?
80. Would the second ?
81. Would the eleventh?
S2. Would the twentieth?
83. Would the twenty-second ?
84. Would the third?
35. Would the twelfth ?
86. Would the twenty-first?
87. Would the sixth ?
88. Would the thirteenth ?
89. Would the seventeenth ?
40. Would the fifth?
41. Would the fourteenth ?
42. Would the twenty-thurd ?
43. Would the eighth ?
44. Would the seventeenth ?
45. Would the twenty-fourth ?
46. Would the eighteenth ?
47. Would the fifteenth ?
48. What would be the denomination of a figure in each
of the above-named places ?
Ans. The denomination of a figure in the fourth place at the left of
the point would be thousands, that of a figure in the seventh place at
the left would be millions, that of the tenth would be billions, &c.
49. Where must a figure be placed to represent trillions ?
An$ In thf* first place of the fifth period, which is the thirteenth plact,
•A the left of the point
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NOTATION AND NUMERATION.
11
Where must a figure be placed to represent
50. MillMiiB?
51. Ten-millk)ii8 ?
52. Units?
53« Ten-tbonsands ?
54. Quadrillions?
14.
55. Thousamls?
56. Hundred-trillions?
57. Hundreds?
58. Hundred-billions?
59. Ten billions?
Method of reading Number*.
Exercises.
(a.) In reading a number represented by figures, we ordiiui*
rily commence at the lefl hand, and read each period as though
its figures stood alone, giving afterwards the name of the period.
For instance, the naraber 42,000,070,294,600,706 would be read in tho
Mine way, and would expre8s,the same value, as if written 42 quadrillions,
70 billions, 294 million^, 600 thousands, and 706.
NoTB. — The scholar should regard a mistake in reading numbers as
one of the most dangerous which can be made, for he will not only bt
likely to copy the numbexs in the same manner as he reads them, but lis
will give those to whom he reads a false idea, which, unless they baTt
the figures before them, they cannot correct
(^.) Bead each of the following numbers : — *
1.
43,271.
16.
607,429.
2.
600,207.
17.
1,579,432.
3.
24,000,217.
18.
914,307,426.
4.
63,279,412.
19.
53,729,415.
5.
432,160,023.
20.
21,437,986,512.
6.
70,000,000.
21.
42,000,042,042.
7.
86,102,102.
22.
547,547,547,547.
8.
160,437,986,216.
23.
101,101,101,101.
9.
20,020,020,020.
24
2,002,002,002,002.
10.
200,280,200,200.
25.
130,201,040,999,999.
11.
70,000,007,700,077.
26.
78,006,200,474.
12.
2,008,002,008,002,008.
27.
900,000,726,000.
18.
6,000,070,000,600,007.
28.
74,206,372,704.
14.
8,200,020,006,307,004.
29.
407,000,000,030,002.
15.
73,052,700,060,007.
30.
703,700,000,000,006.
(c.) Explain the use of the figures used in writing the aboTO
numbers. (See model following the 10th example in 10.)
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13 NOTATION AND NUMERATION
IS* All intervening Places to be JUhd.
(a.) When, as is usually the case, the places are odIj marked
by the figures occupying them, every place between any given
figure and the point must be occupied by some one of the
digits, for otherwise it will be difficult or impossible to tell
the place or denomination of the given figure.
JUustration, The 3 of the number 3 24 may mean 3 thousands, or
3 ten-thousands; but when the space between the 3 and 2 is filled, the
denomination of the 3 is at once apparent Thus, in 3024, or 3124, or
3724, the 3 represents 3 thousands; but in 30024, 31024, 32324, or
37824, it represents 3 ten-thousands.
(3.) The digit to be used in any intervening place is deter-
mined by the number of units to be represented of the denom-
mation of that place. If there are none, then zero should be
used ; if one, then 1 ; if two, then 2 ; &c
Illustration. In order that 9 may represent 9 ten-millions, it must be
written in the eighth place, at the left of the point, and hence there must
be seven figures to the right of it If we wish to express only 9 ten-
millions, or 90 millions, the intervening places must be filled by zeroes,
thus, 90,000,000 ; but if we wish to express 90 millions, 3 thousand, 8
hundred, and 7, we write 9 in the eighth place, as before, 3 in the fourth,
8 in the third, 7 in the first, and zeros in all the intermediate places!
thus, 90,003,807.
16* Metliod of writing Nutnhers^
Examples.
(a.) In writing numbers by figures we may begin at the
left, and write in each successive period the figures necessary to
express the required number of units of the denomination of
that period ; or we may begin at the right, and write in each
successive place the figures expressmg the requirai number of
units of the denomination of that place, taking care to write a
zero in each place otherwise unoccupied.
(6.) In writing numbers be careful to distinguish the decimal point
from the commas used to separate the periods. The former should be a
dot, so carefully made that it cannot be mistaken for a comma, while the
latter should be made with equal care. No number has more than ont
iedmal poiut
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NOVATION AND NUMERATIOX 13
(e.) Atcnracj in reading and writing numbers is of the greatest im«
^rtance. If the numbers used in solving a problem are copied incor*
rectly, the results obtained will of necessity be wrong ; and if the book
from which the problem was obtained is not at hand, or if the facts and
transactions on which the problem was based are forgotten, it will bo
very difficult, and usually impossible, tc make the necessary correo*
tion. <
1. Represent by figures five hundred and twenty-eevea
thousand, four hundred and eighty-nine.
2. Eight thousand, four hundred and seven.
3. Eighty-five thousand.
4. Eighty-five thousand and one.
5. Eighty-five thousand and thirty-one.
6. Nine million, eight hundred and fifty-sis thousand, seven
hundred and twenty-one.
7. Twelve million, twelve thousand, and twelve.
8. Four billion, eight hundred seventy-six million, five hun*
dred and four thousand, three hundred and one.
9. Four billion, eight hundred and four million, eight hun-
dred and four thousand, eight hundred and four.
10. Thirty-seven million, eight hundred and fifty-nine
thousand.
11. Thirty-seven billion, eight hundred and fifly-nine mil-
lion.
12. Thirty-seven billion, eight hundred and fifly-nino
thousand.
13. Thirty-seven million, eight hundred and fifly-nine.
14. Forty billion, three hundred and forty million, four hun-
dred and eighty-seven thousand, &ve hundred and nine.
15. Five biUion, eight hundred and seventy-six thousand,
Beven hundred and forty-six.
16. Seventy-five trillion, eight hundred and seventy-six
billion, four hundred and eighty-two million, four hundred and
seventy-six thousand, three hundred and twenty-seven.
17. Four trillion, seven hundred and sixty-four billion, eight
hundred and twenty-one million, six hundred and seventeen
thousand, four hundred and fifty-one.
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14 NOTATION AND NUMt:RATlOX.
18. Seven hnndred and twenty-five tnllion, eiglit hundred
ind seventy-six billion, four hundred and three million, ciglit
hundred and fifty thoustuid, four hundred.
19. Three hundred and six trillion, eighteen billion, four
hundred million, three thousand, four hundred and seventy-five.
20. Three trillion, three hundred and ninety -nine billion,
three hundred and ninety-nine million, three hundi'ed thousand,
four hundred and three.
21. Eighty-seven trillion, five hundred and four billion, three
hundred million, seven thousand, six hundred and seventy-five,
22. Seventy quadrillion, seven hundred and seven billion,
eeven thousand and seven.
23. Eighty-seven quintillion, eight trillion, seven hundred
billion, eight hundred and seventy thousand, and eighty-seven.
21. Three hundred and fifty-four sextillion, four hundred
and seven quintillion, two hundred and nine quadrillion, nine
hundred and seventeen trillion, seven hundred billion, eighty-
six million, seven thousand, eight hundred and fifty-two.
25. Seven hundred and seven quintillion, two hundred and
six thousand.
26. Five hundred and seventy quadrillion, five hundred
and seventy.
27. Eight sextillion, eight trillion, eight thousand, and eight.
17, Any Combination of Figures may he read as though
alone.
(a,) Combinations of figures, wherever placed, can be read as
though they stood alone, if the name of the place of the right
hand figure be given after reading the figures.
For instance, 347 always stands for, and may be read as, threa
hundred and forty-seven of the denomination of the place occupied by
the 7.
(6.) To illustrate this still further, we have written opposite each of
ihe following n'lmbers the value expressed by 347 in that number.
1. 347,241, ... 347 thousands.
2. 347,100, ... 347 thousands.
3. 4,1 34,7 VI, ... 347 hundreds.
4. 93,476,258,675, . . 347 ten-millioni.
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KOTATION AND NUMF.RATI02«.
15
(c.) Let tlie pupil give the value expressed by the 409 in
each of tlie i'ollowiug numbers, and also the value expressed
by the 27 : —
1.
40,927.
C.
508,340,927.
2.
274,090.
7.
40,927,534.
8.
8,172,7G4,09G,134.
8.
27,409.
4.
52,700,400,273.
9.
4,092,708,375,074.
5.
134,090,0G2,74G.
10.
1,973,409,327,547.
18* Analyns of Numbers.
1. What is the greatest number of tens that can be taken
fr»)m 540,372?
Ans, 54G37 tens.
2. What is the greatest number of ten-thousands that can
be taken from 53,075,423,697 ?
Ans. 5307542 ten-thoustmds.
What is the greatest number of hundreds that can be
taken
3. From 8,643,792 ?
4. From 27,948,073 ?
5. From 2,876?
6. From 487,962?
7. From 79,762?
8. From 279,876,372 ?
9. From 25,986,137,953?
10. From 542,763?
11. What, is the greatest number of thousands that can be
taken from each of the above numbers ? of tens ? of millions? *
•>f billions ? of hundred thousands ? of ten millions ?
12. Express the value of 457869 in hundreds and units.
Ans, 457869 equals 4578 hundreds and 69 units.
Express the value of each of the following in hundreds and
•nits : —
13.
8,796,784.
17.
6,972.
14.
86,724.
18.
79,843.
15.
7,807,375.
19.
987,509,875.
16.
46,739,725,876.
20.
570,072,307,679.
• If any number is less than a million, as 2876, the answer may bi^
3876 u Usa than a million, and therefore no millions can be taken from it.
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21. Express the Taloe of eack of the above m teas amk
mats, lo teo-tbousaiids and units.
22. Express as much of the ralne of each of the abore m
j<m can in ten-tboosands ; as mcch of the remainder as ym
can in hondiedsy and the rest in onits.
Moddi/Amswa', 879G7M = S79 tca-thooaads, €7 hntedi^ aai M
23. Express as nmdi of the Tahie of each of the aboTe aa
joa can in ten-miOions^ as modi of the remainder as joa can
in Aoosands, and the rest in units.
19* Comparisem of the Vabtes represaded hyike sawte^gmn
in different J^aees,
(a.) Since, as we have seen, the figure in an j ph^e repre-
sents ten times the valne it would represent if written one pUu»
£irther towards the right, one hmidred times the vahie it would
represent if written two places farther towards the right, &C.,
it follows Uiat it most represent one * tenth of the yalue it
wonld represent if written one place farther towards the left,
one one-* hundredth of the value it would represent if written
two places farther towards the left, Sec
KoTS. — Tfie expretfion **Kainbers increase firom right to left in a
lenfbld rttio,** if not s tme ttatement of the fact
A wtU of one decinuil denomination always bears the same ratio to a
uml qH the next higher that 1 does to 10 ; hot the ratio which the valae of
a figure of one decimal denomination bears to the ralue of a figvirt of
the next higher is as 1 to 10 onlj when the figures are alike. For in-
stance, in 22 the Talne of the left hand figure is ten times that of the
right tiand figure, while in 25 it is only four times, and in 91 it is ninety
times.
Even when the figures are alike the ratio of increase is not tenfold.
We increase a number by adding to it To increase a number once, one
addition ronst be made to it To increase it twice, two additions must
bo made, &c. A number increased by once itself will produce twice it-
* If the explanations on page 3d are not sufficient to enable the pupil
to understand the meaning and use of the fractional expressions contained
n this section, he should study the first part of the section on fractions
ietme going tarther
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NOTATION AND NCMEHATION. 17
jelf , lucrcnsed by twice itself, will produce three times itself; Vioreased
by nine times itscll', will protluco ten times itself; increased by ten times
itself, will produce eleven times itself, &c. Twenty is ten times two, or
ten times as large as two, but is only nine times larger ; for if we make
two larger by nine times itself, that is, if we add nine twos to two, the
result will be twenty, equal to ten times two.
1. Compare the values expressed by the 4's m the num-
ber 444.
Ans. The first 4 at the right represents one tenth of the valae of
the second 4, and one one-hundredth the value of the third 4 ; the
eeooxd 4 represents ten times the value of the first or right hand, and one
tonth of the value of the third or left hand 4 ; the third 4 expresses ten
times the value of the second, and one hundred times that of the first.
2. Comj)are the values expressed by the 6's in 66 ; in 666 ;
in 606 ; in 660.
3. Compare the values expressed by the 8's in 88 ; in 880 ;
in 808 ; in 8888 ; in 8008.
4. Compare the values expressed by the 3*8 in 333 ; in
3030 ; in 3303 ; in 333333.
30 • Places at Right of Decimal Point*
(a.) In conformity with the same principle, a figure written
in the first place at the right of the point would represent tenths
of units, or simply tenths ; one written in the second place at
the right would represent hundredths of units, or simply hun-
dredths, &c
(5.) Hence, the places at the right of the point have been
named as indi :ated below.
-•I I
3 S §^ g 3g s Si
.000000000
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18
NOTATION AND NUMERATION.
1* Which place from the point is occupied bj the millionthi
Al^re?
Ans, The sixth place at the right
' Which place from th^ point is occupied by the
2. Tenths' figure?
8. Hundi-edths'?
4. Ten-thousandths'?
5. Millionths'?
6. Thousandths'?
7. Billionths'?
8. Hundred-thousandths'?
9. Hundred-millionths* ?
10. What would be the denomination of a figure in the
second place at the right of the point?
11. In the fifth?
12. InthesixOi?
13. In the fourth?
14. In the seventh ?
15. In the first?
16. In the ninth?
17. In the eighth?
18. In the third?
SI* Method of reading Numbers expressed hy Figuren ai
Biglu of Decimal Point.
(a.) Numbers expressed by figures written at the right of
the point are read on exactly the same principles as thoM
expressed by figures at the left.
Thus, if we wish to read .37, we observe that the right hand figure is
in the hundredths* place. We read it then as though written thirty*
feren one hundredths, or •^^. The following will furnish further il«
lustrations of the principle involved : —
1, .86 =x8„v
2. .0086 = T^ATr-
8. .0349 = y§§|„.
4. 8.49 =m =3tV<j-
6. 6.07 =fgj =^^U-
6. 10.1 =-V>^ =10-i\,.
7. 80.03 = ^o'rf'- = 30t§5.
8. 7.008706 = ISSSJSS = Ir^^'^T!.
(i.) Bead the following, expressing the value in all cases
where it is greater than unity, both as an improper fraction
and as a mix ^ number : —
9. .086 = T«5w.
10. .349 = ^3^5^.
11. .000349 = T,,§S?„,
12. 34.9=a^4 = 34^j.
13. 1.01 =1^=1^*^.
14. 300,3 = a^ga = goO-^
15. 3.003 = *8ft3=3TA„.
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VOTATIOK AMD HUSIEnATIOir.
If
1«.
.73
26.
60.05
17.
.08
27,
0.0764
18.
.798
28.
37.9427
19.
6.4
29.
876.5874-
20.
7.0037
80.
8299.4856328
21.
4287
31.
lOOOOOO.OOOOOOl
22.
.940094
82.
87.203794
23.
8.06006
33.
87.000000079
24.
40.7000407
84.
.000000007
25.
21.3804206
85.
2006.000002008
33. Decimal FracHonSy Method of writing them*
(a.) Any fraction whose denominator is a power* often maj
be expressed by writing the numerator so that its right hand
figure shall occupy the place of the same name with its d^
nominator, and omitting the denominator. Fractions thus
written are called Decimal Fractions^ to distinguish them from
Vulgar Fractions, or those whose numerator and denominator
are both written.
NoTB. — The first fifteen examples of 31 are written both as decimal
and vulgar fractions, while the last twenty are onlj written as deeimal
fractions.
The greatest liabilitj to error in writing decimal fractions Is la
placing the point. The learner should bear this in mind| and take pains
to guard against it, remembering that if the point is not correctly
placed) each figure of the number will express a wrong value.
I/'
(6.) Write die following in the form of decimal fractions : —*
1. Eight tenths.
2. Fifty-four hundredths.
8. Eight hundred and seventy-five thousandths,
4. Fifty-four thousandths. '
5. Eight hundred and sixty-four ten-thousandths.
6. Six, and eighty-seven hundredths.
7. Six, and eighty-seven ten-thousandths.
8. Four hundred and thirty-seven tenths.
• For adbut.«B«ftlis wort i»wsr,McaitidbI0ii<4.)
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20
[NOTATION AND NUMERATION.
9 Three hundred and four thousands, and three humlrcd
and four thousandths.
10. Three hundred and four thousands, and three hundred
and four millionths.
(c.) Write the following in the form of decimal fractions : —
11. A.
21.
^
•
31. mo%-
12. -fife.
22.
r/w
S2. jUirr-
13. tSt7-
23.
H-
33. rs^inr.
14. ■A'W.
24.
m-
84. rii§5inT.
15. iM^.
25.
Tl?(^.
85. mm-
16. mi-
26.
f§§iS-
36. TTrfVlftJFTr.
17. 3A.
27.
97t5,j.
37. 500-rl5.
18. 40Ti„.
28.
SOtO,.
88. 50t,At^,-
19. 24t^.
29.
SyfTT-
89. 5y^^jj.
20. 4//,,.
30.
"VtrV
40. 637t^^j,
41. 46948,1^2^%,
48.
i85y6-n?oyw. -r
42. M*M-MSi^-
49.
842X,W,^.
43. 8270000T^«,j%;nr.
50.
40064^8S5^^.
44. aJiW^b'SP.
51.
8^?iLa.
45. IfooVod.
52.
10O00'^OT,n,i;^o.
46. 6000^,^^.
53.
487t^^^'A.
47. 7200000000
^loooSSo
innr*
54.
SSltyVVzy.
Exercises in determining Position of Point.
1. Where must the point be placed in order that the figui^ea
746 maj represent tenths?
Aiu, Between the 4 and 6 ; thns, 74.6
2. Where must the point be placed in order that the figures
746 may represent millionths ?
An8, Six places to the left of the 6 ; which requires that three places
be filled with zeros ; thus, .000746
8. Where must the point be placed in order that the figures
578 may express 573 tenths? hundredths? units? tens? hun-
dreds? thousands? thousandths? millions? millionths?
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NOTATION AND NUMERATION. SI
4. Where must the point be placed in order that the figufM
67064 maj express 87064 hundreds? hundredths? ten«lhoii*
»ands? ten-thousandths? hundred-millions? hundred-mil-
lionths? tens? tenths?
5. Where must the point be placed in order that the figorei
497837 may express 497837 units? millions? biUionths?
hundredths? millionths? hundreds? hundred-thousandths?
24 • Mffect of changing Place of Point,
Multiplication and Division by Fowbrb ov 10.
(a.) Since (8^ 19) each figure of a number represents 10
times the valye which it would represent if written one place far*
ther towards the right, 100 times the value it would represent if
written two places farther towards the right, &c, and ^ of tlie
value it would represent if written one place farther towards
the left, -j^ of the value it would represent if written two
places farther, &c., it follows that removing the figures repre*
senting a number one place farther towards the lefi, or, which
is the same thing, removing the point one place to the rights
multiplies the number by 10, while removing the figures one
place to the rights or the point one place to the /e/?, divides it
by 10. A change of two places would in like manner multi-
ply or divide by 100, of three places by 1000, &c
These principles generalized would be stated thus : — •
(b.) To express the product of any number multiplied by any
power of 10, remove the decimal point as many places to the
right as there are zeros used in writing the given multiplier,
(c.) To express the quotient of a number divided by any
power of 10, reinove the decimal point as many places to the
left as there are zeros used in writing the given divisor,
(d,) When, by such change, any places between the number
midpoint are left vacant, they must be filled by zeros.
Note. — The rule usually given for multiplying by the powers of 10,
li to " annex as many zeros to the multiplicand as there are zeros in the
multiplier." It is, however, a very defective one, as it only applies wImq
the decimal point is not written, and then only multiplies the number by
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22
NOTATION AND NUMKRATION.
changing the place to which we refer the point It is the more cbjoc
tionable, ns it tends to convey the false idea that the zero is essential to
tlic multiplication. As a matter of fact, annexing any figure whatever
to a number will, if the decimal point is omitted, multiply It by 10, for
it will change the place to which the decimal point is referred ; but if
the annexed figure is other than zero, its value will be added to the prod-
act of the number by 10. Thus, annexing 7 to 43 gives 437, which
equals 10 times 43, plus 7.
The principle involved is this : that every change which is made in the
position of figures with reference to the decimal point, — whether it is
made by changing the position of the point, or by writing other figures
between the given figures and the point, — alters the value they repre-
sent, by multiplying or dividing them by 10 or some power of ten. A
figure can only alter the value expressed by other figures when it i
written between them and the point.
in fi<;ures the results of the fol
(e.) How will you express
lowing indicated operations ?
1. 87X10*
2. .87 X 10
8. .0087 X 10
4. 87-7-10
5. .087 -MO
6. 870 -r- 10
7. 5.7 -T- 100
8. .057-7-100
9. .8278 X 100 '
10. 4.5786 X 1000
11. .4-7-10
12. 479.643 X 1000
13. 479.643-7-1000
(/.) Let the student now tell by inspection, without ctiai>*
ging the place of the point or re- writing the numT»er3, tne re-
sult of the above indicated operations.
• The student should remember that when the d^imai point is not
giarlced, it is always uuders^ ->od to belong at the right of the given figures.
14.
4279 X 1000
15.
.6307 X 100
16.
.00694 X 10
17.
5429 -7- 100
18.
400.794-7-10
19.
.0054279 -7- 1C)000
20.
.0004 X lOuOOOO
21.
.002 X lOuoO
22.
348.796 X lOa
23.
2 X 1000000
24.
2 ^ 1000000
25.
.006 X lOoOOOOOOO
26.
.006 ■- lOOOCODOOO
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NOTATION AND NUMERATION. 23
Thus . eighth-seven mMjtfied by ten equah eight hundred and seventy ;
eightg'Uven hundredths multiplied by ten equals eighty-seven tenths^ or eight
and seven tentlis^ &c. He bhoold learn to do this without the slightest
hesitation.
35 Change of Denomination.
(a.) The value of a number may be expressed in terms of
any other decimar denomination as well &s in units, by making
the requisite change in the place of the point. Thus : —
847 =s 34.7 tens, = 8.47 hundreds, = .847 of a thousand, ss .0847 of
A ten-thousandf &c.
847 = 8470 tenths, = 84700 hundredths, = 847000 thousandths, &c
642.06 = 6.420G hundreds, = 6420G hundredths, &c
(b.) Express the value of each of the following in tenths,
then in tens ; in hundredths, then in hundreds ; in miUionths,
and then in millions : —
1.
4327
4. 2700
7. 4683.7642
2.
82794.6
5. .048
8. .00006
3.
.437
6. 2.7
9. 8000000
SO. French Method of Numeration.
The foregoing method of numeration is called the French
method. It divides the figures expressing a number into
periods of three figures each, making a unit of one period
equal to one thousand units of the next lower period.
Thus one million equals one thousand thousands \ one billion equals
one thousand millions \ ,one trillion equals one thousand billions, &c
37« English Method of Numeration.
(a.) There is another method, called the English method^
which divides the figures expressing a number into periods of
six figures each, making a unit of one period equal to ouo
million units of the next lower period.
{b.) By this method one billion equals one million millions ;
one trillion equals one million billions, &c. This is illustrated
in the following example : —
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34 NOTATION AND NUMERATIOM.
lili
9,7 5 2 730,605897,2530 60,9 345 78.
s*l ii 5?
'I
(c.) In this method, as in the French, we read the figures
in each period as though thej stood alone, calling afterwards
<he name of the period.
{d,) The above number would be read, 9 quadrillions, 752730 tril-
lions, 665897 billions, 253060 millions, 934578.
38. Comparison of French and English Methods,
(a.) It will be readily seen that while the English periods
bear the same name as the French, and while one thousand
and one million represent the same number in the two sys-
temS; one billion, one trillion, or a unit of any higher denomina-
tion is much greater in the English system than in the French.
Thus, an English billion equals a French trillion ; an English trillion
equals a French quintillion.
(6.) The French method is the one generally used in this country and
on the continent of Europe, and being much more convenient than the
English, has been adopted in part in England, and is likely to come into
general use there.
Sn. Numbers to he read according to the English Methods
1. 426,794798,764387.
2. 86432,795876,942759.
«. 287000,568975,006723.
4. . 6,456327,309670,800659. .
5. 825,897563,475003,900065.654008-
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NOTATION AND- NUMERATION. 25
30. The Roman Method.
(a.) The Roman method of notation represents numbers by
letters of the alphabet.
(6.) It is now chiefly used in numbering the sections or chapters of a
b-)ok, the pages of a preface or introduction, the year of the Christian
era, or when it is necessary to distinguish one class of numbers from
aaother.
(c.) The letters used are the following, viz. : —
I , which stands for One.
V , which stands for Five.
X, which stands for Ten.
L , which stands for Fifty.
C , which stands for One Hundred.
D , which stands for Five Hundred.
M , which stands for One Thousand.
frf.) Other numbers are represented by repetitions and
combinations of these letters.
(c.) Wlieu a letter is repeated, it indicates that the number
it represents is to be repeated.
Thus: II. = two J m. = three ; XX. = twenty j XXX. = thirty, &c.
(/.) If a letter expressing one number be placed before a
letter expressing a larger number, the former is to be sub-
tracted from the latter; but if the letter expressing the larger
value be placed first, the values of the two are to be added
together.
Thus ; IV. = four ; IX. = nine ; XL. = forty, &c. ; while VI. — six
XI. s=s eleven j LX = sixty, &c.
(^.) In the following columns, the letters stand for the
niunbers written against them : —
I. . One.
VII. . Seven.
n. . Two.
Vin . Eight.
m. . Three.
IX. . Nme.
IV. . Four.
X. . Ten.
V Five.
XL . Eleven.
VL . Six.
XIL . Twelve.
3
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S6
NOTATION AND SUMliKATIOS.
XIII.
. Thirteen.
L.
.Fifly
XIV.
. Fourteen.
LX.
. Sixty.
XV.
. Fifteen.
LXX.
. Seventy.
XVL
. Sixteen.
LXXX.
. Eighty.
XVII.
. Seventeen.
xc.
. Ninety.
xvin.
. Eighteen.
XCIX.
. Ninety-nine.
XIX,
, Nineteen.
c.
• One Hundred.
XX,
. Twenty.
CL.
. One Hundred luid
XXL
. Twenty-one.
Fifty.
xxu.
. Twenty-two.
CLXXXVin. . One Hundred
XXIII.
. Twenty-three.
and Eighty-eiglit
XXIV.,
Twenty-four.
CC.
. Two Hundred.
XXV.
. Twenty-five.
CCC.
. Three Hundred.
XXVI.
. Twenty-six.
CD.
. Four Hundred.
XXVII.
. Twenty-seven.
D.
. Five Hundred.
XXVIII.
. Twenty-eight.
DC.
. Six Hundred.
XXIX.
, Twenty-nine.
DCC.
. Seven Hundred.
XXX.
. Thirty.
DCCC.
. Eight Hundred.
XXXI.
, Thirty-one.
CM.
. Nine Hundred.
XL.
. Forty.
M.
. One Thousand.
XLL
• Forty-one.
MDCCCLIV. . 1854.
XLIX.
• Forty-nine.
(h.) A dash placed over a letter makes it express t^iousands
instead of ones. Thus, VTzr: 5000; VL= 6000 ;"L; = 50,000;
SC.= 90,000, &c
(u) Bead the following numbers : —
XCVIII.
DCCXLII.
MDCCLXXVI.
XXDCCCXCIX.
CCXLVIIICCCL.
DCLIDLXX L
DCCCXCIXC CCXXXUL
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TABLKS OF MONEY, WEIGHTS, AND MEASURES. 27
SECTION III.
TABLES OF MONEY, WEIGHTS, AND MEASURES.
31* Introductory.
{n.) All nations, excepting perhaps the most barbarous, have
some kind of money ; but each nation has a system peculiar
to itself, aivl generally diffenng from every other in its de-
nominations, coins, &c. The people of the United States
reckon money in dollars and cents, the English reckon it in
pounds, shillings, and pence, and the French in francs and
centimes.
(h.) So too each nation has a peculiar system of weights and
measures, some of the most important of which are illustrated
in the following tables.
32. United States, or Federcd Money,
(a.) The money of the United States is called Federal
Money.
TABLE OP FEDEBAL MONET.
10 mills = 1 cent.
10 cents = 1 dime.
10 dimes = 1 dollar.
10 dollars = 1 eagle.
[b.) The coins of the United States are the dollar, the half
dollar or fifty-cent piece, the quarter dollar or twenty-five-cent
piece, the dime or ten-cent piece, the half dime or five-cent
piece, the three-cent piece, the cent ; the eagle or ten dollar
piece, the double eagle or twenty-dollar piece, the half eagle
or five-dollar piece, and the quarter eagle, worth two and a
half dollars.
(c.) The dollar is coined both of gold and silver ; the coins
worth more than a dollar are of gold, and the others, with the
exception of the cent, are of silve-. The cent is of copper.
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28 TABLES OP MONEY, WEIGHTS, AND MEASURES.
(d.) Values in federal money are usually expressed in dol-
lars and cents, or in dollars, cents, and mills, the dollar being
regarded as the unit, the cent as one hundredth of it, and the
mill as one tenth of the cent, or one thousandth of the dollar.
Indeed, dimes, cents, and mills being nothing more than tenths,
hundredths, and thousandths of a dollar, the figures represent-
ing them may be, and usually are, written in the form of a
decimal fraction, and may with perfect propriety be read as
such.
(e.) The character $, placed at the left of figures, shows
that they stand for dollars.
Illustrations, (a.) $3.75 = * 3 dollars and 75 cents, = 3 dollars and
jyjj. of a dollar, = 3 dollars, 7 dimes, and 5 cents, &c.
(6.) $237^64 = * 237 dollars, 26 cents, and 4 mills, = 237 dollars
and ^j3^ of a dollar, = ^f J§g^ of a dollar, = 23726.4 cents, =»
237264 mills, &c.
(c.) $7,042 = * 7 dollars, 4 cents, and 2 mills, = 7 dollars, 42 mills,
es 7 dollars and * 2 of a dollar, = |.g^^ of a dollar, = 7042 mills, =
704.2 cents, &c.
Exercises.
(/.) Read each of the following, giving the value 1st in dol-
lars and cents, or, where there are mills, in dollars, cents, and
mills ; 2d, in dollai*s and decimal parts c£ a dollar ; 3d, in
ccr
lis ; tLii, lu luii
ds;— •
/.TV
/ iTv-x;
1.
$ 6.79
7.
$ .073
13.
$ 7084.79
2.
$ 8.03
8.
$ 30.07
14.
$ 400.06
3.
$ 764.37
9.
$ 2587.00
15.
$ .125
4.
$ 28.00
10.
$ 25.00
16.
$ 97.886
5.
$ 5.976
11.
$ 19.875
17.
$ 20.07
6.
$ .073
12.
$ .625
18.
$ .06
33« English Money,
(a.) The money used in England is called English, or Ster-
ling Money.
* The form marked with a star represents the usual method of readizig
these Talues.
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TABLES OF MONEY, WCIOUTS, XND MEASUTIBS. 29
TABLE OF STERLING MONET.
4 farthings = 1 penny.
12 pence = 1 shilling.
20 shillings = 1 pound.
{fi.) When numbers expressing values in sterling money
are used, the character £ marks the pounds, s. the shillings,
d. the pence, and qr. the farthings.
Illustration. 27 pounds, 18 shillings, 5 pence, and 3 farthings, may
£ t. d. qrt.
be written thus : 28£, 18 s. 5 d. 3 qrs. ; or thus : 27 18 5 3. Farthings
are also frequently expressed as fourths of a penny. Thus, 7 d. 3 qrs. =3
7M., and both forms may be read as 7 pence and 3 farthings.
(c.) The coin representing the pound is called the sover-
eign. Its value in United States money varies from four
dollars and eighty-three cents to four dollars and eighty-six
cents, but is usually about four dollai*s and eighty-four cents.
There are several other coins used in England, of which we
will mention only two, viz., the guinea, or twenty-one shilling
piece, and the crown, or five-shilling piece.
34. Avoirdupois Weight.
(a.) Almost all articles, except gold, silver, and jewels, are
weighed by what is called Avoirdupois Weight.
TABLE CF AVOIRDUPOIS WEIGHT.
16 drams = 1 ounce.
16 ounces = 1 pound.
25 pounds = 1 quarter.
4 quarters = 1 hundred weight.
20 hundred weight = 1 ton.
(ft.) The abbreviations made use of in this weight are
T. for ton, cwt for hundred weight, qr. for quarter, lb. for
pound, oz. for ounce, and dr. for dram.
T. ewt qr. lb. os. dr.
Forexample, 5T.l7cwt.3qr. 13lb. Soz. 7dr. = 5 17 3 13 8 7=«
S tons, 17 hundred weight, 3 quarters, 13 pounds, 8 ounces, and 7 drams.
(e.) Formerly the quarter was reckoned at 28 pounds, the hundred
r
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30 TABLES OP MONET, WEIGHTS, AND MEASURES.
weight at 112 pounds, and the ton at 2240 pounds, and they are so
reckoned at the present time in Great Britain as well as in the standard
of the United States government. Most of the states of the Union
have, however, passed laws fixing the values as in the table, and they
are almost always so reckoned by merchants in baying and selling.
35. Troy Weight,
(a.) The weighty used in weighing gold, silver, and precious
stones is called Troy Weight. This weight is also used ill
philosophical experiments.
TABLE OP TROT WEIGHT.
24 grains = 1 pennyweight.
20 pennyweights = 1 ounce.
12 ounces = 1 pound.
(ft.) In this weight, lb. stands for pound, oz. for ounce,
dwt for pennyweight, and gr. for grain.
Illustration. 13 pounds, 7 ounces, 18 pennyweights, 23 grains, may
lb. oz. dwt. gr.
be expressed thus : 13 lb. 7oz. 18 dwt. 23 gr.; or thus: 13 7 18 23.
(c.) The "carat," which equals four grains, is used in weighing dia^
monds. The term carat is also used in stating the fineness of gold, and
means the twenty-fourth part of any weight of gold or gold alloy. Pure
gold is " 24 carats fine." Gold is 22 carats fine when ^2 of it is pure
gold and JL is alloy.
3G. Apothecaries^ Weight.
(a .) The weight used in compounding or mixing medicines is
called Apothecaries' Weight. Physicians write their prescrip-
tions in this weight, but medicines are bought and sold by
AToirdupois Weight.
^ TABLE OF apothecaries' WEIGHT.
20 grains = 1 scruple.
3 scruples = 1 dram.
8 drams = 1 ounce.
12 ounces = 1 pound.
To mark the denominations of this weight, we use the fol-
lowing characters, viz. : ft) for pound, | for ouqce, 3 for dram,
*^ for scruple, and gr. for grain,
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TABLES OF MONET, WEIGHTS, AND MEASimEt). 51
For example: 51^63 43 2^ 17 grs^ would be read as 5 poandSi
6 oonces, 4 drains, 2 scruples, and 17 -grains.
37. Comparison of Troy, Avoirdupois^ and Apothecaries*
Weights,
[a.) The only difference between Troy Weiglit and Apoth-
ecaries' Weight is, that in the former the ounce is divided into
pennyweights and grains, while in the latter it is divided into
di*ams, scruples, and grains. The pound, ounce, and grain are
the same in both weights.
(h.) The value of denominations of the same name in
Avoirdupois' and Troy Weights differs very materially, as
aiay be seen from the following table, which shows the valuo
sn Troy grains of each denomination we have given in tho
preceding tables of weights.
TABLE OP COMPARISON.
1 lb. Av. . =r 7000 gr. Troy.
1 lb. Tr. =
lib
= 5760
«
u
1 oz.Av.
= 437^
((
a
1 oz. Tr. =
IS
= 480
((
u
1 dr. Av.
= 27i,i
((
u
1 5
= 60
«
u
1 B
= 20
((
u
1 dwt.
= 24
u
u
1 gr. Ap.
= 1
u
a
(c) From the above table we should find by calculation
that
144 lb. Av. = 175 lb. Tr.,
and 192 oz. Av. = 175 oz. Tr.
Tlierefore, 1 lb. Av. = iJJ or 1^3^ lb. T%,
and 1 oz. Av. z= ^^J J oz. Tr.
(d,) Which is the heavier, and why, —
1. A pound of gold or a pound of feathers^
2. A pound of lead or a pound of feathers ?
3. A pound of gold or a pound of lead ?
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82 TABLES OF MOXEY, WKIGHTS, AND MEASURES*
4. An ounce of gold or an ounce of feathers ?
5. An ounce of lead or an ounce of feathers ?
6. An ounce of lead or an ounce of gpld ?
38 • Long Measure,
{a.) Distances in any direction are measured bj Long
Measure.
TABLE OF LONG MEASURE
12 lines = 1 inch.
12 inches = 1 foot.
3 feet =z 1 yai-d.
^ y Z ' r = 1 rod or pole
16^ feet I ^
40 rods == 1 furlong.
8 furlongs = 1 mile.
3 miles = 1 league.
(6.) In this measure, le. stands for league, ra. for mile, fur.
for furlong, rd. for rod, yd. for yard, ft. for foot, and in. fo:
inches.
(c.) Surveyors usually measure distances by means of a
chain 4 rods in length, called Gunter's chain, ci the Survey-
or's chain. This chain contains 100 equal links ; 25 links
will, therefore, equal 1 I'od, and 1 link will equal 7JJ inches.
39. Cloth Measure.
(a.) This measure is used for measuring cloth, ribUons, Sec
TABLE OP CLOTH MEASURE.
\ 2J inches = 1 nail.
4 nails = 1 quarter.
4 quarters = 1 yard.
(h ) In this measure, yd. stands for yard, qr. {or quarter,
and na. for nail.
(c.) The yard and inch ai*e the same in lengtli as the yard
and inai z Long Measure
//^Wr-^ /; ^ir :i^jA^
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TABLES OF MONEY, WEIGHTS, AND MEASURES.
3a
• 4©. Square Measure.
(a.) Square Measure is used for measuring surfaces.
As true ideas of the nature of the right angle, rectangle, and square
are essential to a just appreciation of this measure, we insert the follow-
ing definitions and illustrations, leaving it with the teacher to give such
others as he maj deem necessary.
(b,) When tyvo lines meet at a point, their difference in
direction is called the angle of the two lines. The point
where tJiey meet is called the vertex of the angle.
FlQ. 1.
(c.) In this figure, the lines marked 6 A
and B C form an angle whose vertex is at B.
In reading an angle, the letter at the vertei.
is always made the middle one. The angle in
Fig. 1 may he read either as the angle ABC,
or as the angle C B A.
(rf.) In this figure there are two an-
gles, viz., D B A and A B C. The
first is the difference in direction of the
two lines A B and D B, and the second
is the difference in direction of the two
lines A B and B C. It is evident that
the first angle is larger than the second.
(e,) When the two angles formed by one straight line
meeting another are equal to each other, they are called right
angles, and the two lines are said to be perpendicular to each
other
^^^'^- (/:) Suppose that the straight line D B
should so meet the straijrht line A C as to make
the adjacent angles A B D and D B C equal
to each other ; then A B D and D B C will
each of them be right angles, and D B and
A C will be perpendicular to each other.
(y.) An angle greater than a right angle is called an obtuse
(blunt) anghy and one lejs than a right angle is called an
acute (sharp) angle.
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di TABLES OF 3I0NEY, WEIGHTS, ANX> MBASLRES.
In Fig. 2, D B A is an obtasc angle, and A B C i« ua acute angli).
In Fig. 1, A B C is an acute angle.
(k.) A four-sided figure, having all its angles riglit angles,
is called a rectangle,
(i') A rectangle, having all its sides equal, is called a
square. A square, then, has four equal sides and four equal
angles.
Figs. 4 and 5 represent rectangles. Fig. 5 also represents a sqnare.
Fig. 4. Fig. .5.
(j.) If a square measures a foot on each side, it is called a
square foot ; if it measures a yard on a side, it is called a
square yard, dec A sqnare unit, or superficial unit, then, is
any surface equivalent to a square 1 unit long and 1 unit wide.
(k.) All surfaces are measured hy Square Measure, that is,
by the number of squares of a given size to which they are
equivalent. Thus, a surface contains 5 square fee', when it is
equivalent to 5 squares, each measuring 1 foot on a side.
TABLE OF SQUARE MEASURE.
144 square inches = 1 square foot.
square feet = 1 square yard.
301 square yards, or ) - ,
272Uuarefeet. J = 1 «q«a^ 'od.
40 square rods = 1 rood.
4 roods = 1 acre.
640 acres = 1 mile.
{h\ In this measure, Sq. M. stands for sqnare mile, A. for
acre, K, for rood, sq. rd. for square rod, sq. yd. for square yard,
0q. ft. for square foot, and sq. in. for square inch.
JMoTB. — Sin:e a surface a unit long and a unit wide contains a square
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TABLES OF MONET, WEIGHTS, AND MEABLRKS. 85
anit, a sniface two. units long and one nnit wide mast contain two square
5inits ; a suifacc three units long and one nnit wide must contain three
iqnare units ; and generally, a surface one unit wide must contain ai
many square units as there arc units in length.
A surface two units wide must contain twice as many square unit3 at
a surface one unit wide, i. e., twice as many as there are units in length;
a surface three units wide must contain three times as many square unitt
as a surface one unit wide, i. e., three times as many as there are units
in length ; and generally, any surface must contain as many square units
as there are in the product obtained by multiplying its length by its
breadth. The surfaces here spoken of are, in all cases, supposed to be
rectangular ones.
41* Cubic Measure,
Cubic Measure is used in measuring solids.
(a.) A solid is a magnitude whicli has length, breadth, and
thickness.
Note. — The term " solid," as used in mathematics, refers to space
lather than to material substances.
(h.) A cube is a rectangular solid, whose length, breadth,
and height are equal. It may also be defined as a solid ivhich
is bounded by six equal squares.
(c.) A cube 1 foot long, 1 foot wide, and 1 foot high would
be a cubic foot. A cube 1 yard long, 1 yard wide, and 1 yard
high would be a cubic yard. A cubic unit, then, is any solid
equivalent to a cube 1 unit long, 1 unit wide, and 1 unit high.
(cL) The solid contents of bodies are measured by cubic
measure, i. e., by the number of cubes of a given size which
the bodies contain, or to which they are equivalent.
TABLE OP CUBIC MEASURE.
1728 cubic inches = 1 cubic foot.
27 cubic feet = 1 cubic yard.
16 cubic feet = 1 cord foot.
8 cord feet, or > , _ ^
<»>o u» /• X r = Icoidofwood.
128 cubic feet, )
(e.) In this measure, C. stands fbr cord, Cd. ft. for cord
fi)ot, cu. ft. for cubic foot, cu. yd. for cubic yard, and cu. in. for
cubic inch.
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86 TABLES OF MONET, WEIGHTS, AND MEASURIJS.
Note. — Since a solid one nnit long, one nnit wide, and one unit
hAf*\\ contains a cubic unit, a solid one unit wide, one nnit high, and two
units long must contain two cubic units ; one tliree units long must con-
tain three cubic units ; one four units long must contain four cubic nnits ;
and generally, a solid one unit wide and one unit high must contain as
many cubic units as there are linear nnits in its length.
Again. Since a solid two units wide must contain twice as many
cubic units as a solid one unit wide, and a solid three units wide miLst
contain three times as many as a solid one unit wide, &C., it follows that
a solid one unit high must contain as many cubic nnits as there are in
the product of its length by its breadth, L e., as many cubic nnits as
there are superficial units in its base.
Again. Since a solid two units high contains twice as many cubic
nnits as a solid one unit high, and a solid three units high contains three
times as many cubic units as a solid one unit high, it follows that any solid
must contain as many cubic units as there are in the product obtained
by multiplying the number of superficial units in its base by the number
of linear units in its height, i. e., as many cubic units as there are in the
product of its length multiplied by its breadth, multiplied by its height
4:i8. Circular or Angular Measure,
(a.) Circular or Angular Measure is used to measure
angles, and the circumferences of circles.
(6.) A circle is a surface bounded by a curved line which
is every where equally distant from a point within called the
centre. The boundary line is called the circumference of the
circle.
Fig. 6 represents a circle of which C is the centre.
(c.) The distance from the centre of a
circle to the circumference is called the
radius of the circle.
{d.) The distance from a point on one
side of a circle thiough the centre to a
point on the opposite side is called the
diameter of the circle.
(e.) Any portion of the circumference is called a^ arc.
(/I) Every circumference of a circle, whether the circle be
■"vge or small, is supposed to be divided into 360 equal pArts^
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TABLES OP MONEY, WEIGHTS, AND MEA8UBE8. 37
called degrees. Each degree is divided into 60 equal parts,
called minutes, and each minute into 60 equal parts, called
seconds.
(g.) A degree is to be regarded simply as the S60th part
of the circumference of the circle considered. Honce, it is
obvious that its length, and that of its subdivisions, must vary
with the size of the circle.
(h.) If from the vertex of an angle, as a centre, we should
dra^r a circumference, some portion of that circumference
would be included between the sides of the angle. The larger
the angle is, the larger will be the arc included between its
sides, and the smaller the angle is, the smaller will be the
included arc.
(t.) Since the angle and the arc thus vary with each other,
the arc is taken as the measure of the angle. If the arc con-
tains 50 degrees, the angle is one of 50 degrees, &c
Fig. 7 illustrates this.
Fig. 7.
(k.) In this figure let C be the centre of the
circle and the vertex of the several angles, and
let F H and G D be lines perpendicular to each
other, and C E and C I be lines drawn at random.
Then the arc E D is the measure of the angle
E C D ; the arc D F, of the angle D C F ; the
arc D I, of the angle D C I ; the arc E F, of the
angle E C F, &c.
(?.) Since F H and G D are perpendicular to each other, the angles
F C D, D C H, H C G, and G C F are all right angles, and each of them
must include i. of the angular space about the point C. Therefore the arc
included between the sides of a right angle equals i of the circumfer-
ence, and the measure of a right angle is J. of 360 degrees, which eqnali
90 degrees. The measure of an acute angle is less than 90 degrees, and
that of an obtuse angle more than 90 degrees.
TABLE OP CIRCULAR OR ANGULAR MEASURE.
60 seconds = 1 minute.
60 minutes = 1 degree.
360 degrees = 1 circumference.
(m.) Degrees are marked bj the character **, minutes by ',
4
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88 TABLES OF MONET, WEIGHTS, AND MEASURES.
seconds by": thus, 13^ 27- 40"=: 13 degrees, 27 minuteij
and 49 seconds.
An are of 90° is called a quadrant
'A sign i? an astronomical measure of 30**.
43. 2)ry Measure,
(a.) Dry Measure is used for measuring all kinds of grain,
beans, nuts, salt, &c.
TABLE OF DRY MEASURE.
2 pints = 1 quart
8 quarts = 1 peck.
4 pecks = 1 bushel.
(b,) The chaldron of 36 .bushels is sometimes used in
measuring coals, ch. stands for chaldron, bu. for bushel, pk.
for peck, qt. for quart, and pt. for pint.
(c.) The bushel contains 2150f cubic inches. The quart
must, therefore, contain 67-J cubic inches.
44. Liquid Measure.
(a.) All kinds of liquids are measured by Liquid Measure.
LIQUID MEASURE.
4 gills = 1 pint.
2 pints = 1 quart.
4 quarts = 1 gallon.
(5.) In this measure gal. stands for gallon, qt. for quart, pt.
lor pint, and gi. for gill.
(c.) The hogshead of 63 gallons is used in estimating the
contents of reservoirs or other large bodies of water ; but in
all other cases, the term hogshead is not a definite measure.
Casks containing from 50 or 60 to 100 or 200 gallons are
called hogsheads.
(d.) A barrel of cider is usually reckoned at 31 J gallons.
(«.) The gallon contains 231 cubic inches.
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TABLES OP MONEY, WEIGHTS, AND MEASURES. 89
(y!) The beer gallon is sometimes used for measuring milk,
beer, and ale. It contains 282 cubic inches. The beer quart
must therefore contain 70^ cubic inches.
45. Comparison of Dnj^ Liquid^ and Beer Measures.
Tlie following table will be convenient for use in comparing
Pry, Liquid, and Beer Measures : —
1 qt. Dry Measure = 67-J cubic inches.
1 qt. Liquid Measure =. 57J cubic inches.
1 qt. Beer Measure = 70^ cubic inches.
46. TABLE OP TIME.
60 seconds = 1 minute.
60 minutes = 1 hour.
24 hours = 1 day.
7 days = 1 week.
8 65 J days, or
52 weeks and \^ days J ^
(a.) The year is divided into 12 months, which differ
Bomewhat in length. In this measure yr. stands for year, mo.
for month, wk. for week, da. for day, h. for hour, m. for
minute, and sec. for second.
(6.) To avoid the inconvenience of reckoning ^ of a day
with each year, every fourth year (called leap year) is reck-
oned at 366 days, and the others are reckoned at 365 days.
A leap year may always be known by this, viz. : Its number
can be divided by 4. Thus we know that 1852 is a leap
year, because 1852 can be divided by 4 without a remainder.
(c.) The year in reality contains but 365 days, 5 h. 48 m.
48 sec ; so that by reckoning 365^ davs we make a slight error
each year, which in 100 years amounts to about 1 day. The
centennial years are not, therefore, reckoned as leap years,
unless the number of the year be divisible by 400. Thus the
year 1900 will not be a leap year ; but the year 2000 will be.
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40 NOTATION AND NUMERATIOK.
TABLE OP THE MONTHS.
January has 31 days.
February * has 28 days.
March has 31 days.
April has 30 days.
May has 31 days.
June has 30 days.
July has 31 days.
August has 31 days.
September has 30 days*
October has 81 days.
November has 30 days.
December has 31 days.
47. Miscellaneous.
12 things = 1 dozen.
12 dozen = 1 gross. /
1 2 gross = 1 great gross.
20 things = 1 score.
A barrel of beef or pork weighs 200 lbs.
A barrel of flour weighs 196 lbs.
PAPER.
24 sheets = 1 quire.
20 quires = 1 ream.
BOOKS.
A sheet folded into 2 leaves is called a folio.
« « « '< 4 " " « " quarto.
u u u « 8 « « « « octavo.
« « " « 12 « « « « duodecimo or 12mo.
it u a u 18 « « « « ISino.
4:8. French Measures and Weights,
{a.) The following measures and weights are often referred to in thif
country, especially in scientific works. .
FBENCH LONG MEASURE.
10 millimetres = 1 centimetre
10 centimetres = 1 decimetre.
10 decimetres = 1 metre.
10 metres = 1 decametre.
10 decametres = 1 hectometre.
♦ In leap year February has 29 days.
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ADDITION. 41
10 hectometres es 1 kilometre.
10 kilometres i» 1 myriametre.
Tlie m«tre is regarded as the anit of measure, and equals 89.371 at
oar inches. It is the twentj-roiilionth part of the distance measured oa
a meridian, from one pole to the other.
(6.) FREKCH WEIOBTS.
10 milligrammes sa i centigramme.
10 centigrammes bi i decigpvmme.
10 decigrammes bi i gramme.
10 grammes ss i decagramme.
10 decagrammes s= 1 hectogramme.
10 hectogrammes =» 1 kilogramme.
10 kilogrammes s= i myriagramme.
The gramme is regarded as the unit of this weight, and eqoab 15.434
Troy grains.
The kilogramme is the weight most frequently used in business trans-
actions, and equals 15434 Troy grains, or very nearly 2.204857 pounds
Avoirdupois.
(C.) FRENCH IIONET.
10 centimes = 1 decime.
10 decimes «= 1 franc.
The franc =3 $.186 ; hence, the five-franc piece, often seen in the
United States, is equal in value to 93 cents.
SECTION IV.
ADDITION.
49. DefinitionSj Illustrations, and Explanations.
(rt.) Addition is the process by which, having
l^eVERAL NUMBERS GIVEN, TVE FIND A NUMBER EQUAL IN
VALUE TO ALL OP THEM.
{h.) The number thus obtained is called the sum Of
AMOUNT.
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42 ADDITION.
Thus : How many are 7 -|- 3 -f- 5, would bo a problem in addition,
And the answer, 15, would be the sum of 7, 3, and 5.
(c.) In order that numbers may be added, it is necessary
that the things they represent shall be of the same name or
denomination.
Illustrations. — 2 books and 3 slates would be neither 5 books nor
6 slates ; but since books and slates are both things, we can change the
denomination of both by calling them things ; when we shall have, 2
things and 3 things are 5 things.
In like manner 2 tens and 3 units would be neither 5 tens nor 5
units ; but by reducing the tens to units, calling them 20 units, we shall
have 2 tens -f" 3 units = 20 units -f- 3 units = 23 units.
2 shillings -|- 3 pence arc neither 5 shillings nor 5 pence; but since 2
shillings = 24 pence, 2 shillings 4~ 3 pence must equal 24 pence 4~ 3
pence, or 27 pence.
(d,) For such reasons it will be found convenient, in writing
large nunibers for addition, to write those of the same denom-
ination near each other. This can best be done by writing
them in vertical columns, so that units shall come under units,
tens under tens, &c., and pounds under pounds, shillings under
shillings, pence under pence, &c.
(e,) In adding, we can begin with any denominatioii we
choose ; but it will usually be more convenient to begin with the
lowest, or the one at the right hand, and to reduce the sum of
each column to a higher denomination, when it can be done.
Note. — Addition is the most important of the four numerical opera-
tions, both because it is the foundation of all the others, and because it is
the one most frequently used in all the departments of practical life.
Moreover, it is the one in which there is the greatest liability to error.
For these reasons, and many others which might be urged, the student
should be very careful to master it fully.
(f,) The methods of applying these principles are illustrat-
ed in the following examples and solutions.
5©. Simple Addition,
(a.) Abstract numbers, or concrete numbers, which repre-
Bcnt values in terms of a single denomination, as in pounds^
in hushehy or in dollars, are called simple numbers ; but
concrete n-imbers, which represent values in terms of sevei'jU
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ADDITION. 43
different denominations, as in pounds, shillings, and pence^ or
in bushels, pecks, and quarts, are called compound nu3IBEKS
(b,) Simple Addition is the addition of simple numbers
r What is the sum of 75798 + 24687 + 39764 + 86328 +
4395 + 283 + 86536 ?
Sdution. — We first write the numbers, placing units under nnita,
tens under tens, &c., in order that figures expressing the same denony
nations may be near together Thus : —
75798.
24687.
39764.
86328.
4395.
283.
86536.
317,791.
* Beginning at the bottom of the units column, (because the lowest
denomination mentioned is units,) and naming only results, we add thus :
6, 9, 14, 22, 26, 33, 41 unit^, which are equal to 4 tens and 1 unit.
"Writing 1 as the units' figure of the amount, we add the 4 tens with
the figures of the tens column ; thus, 4, 7, 15, 24, 26, 32, 40, 49 tens,
which arc equal to 4 hundreds and 9 tens.
Writing 9 as the tens' figure of the amount, we add the 4 hundreds
with the figures of the hundreds column, thus; 4, 9, 11, 14, 17, 24, 30,
37 hundreds, which are equal to 3 thousands and 7 hundreds.
Writing 7 as the hundreds' figure of the amount, we add the 3 thou«
sands with the figures of the thousands column ; thus, 3, 9, 13, 19, 28,
32, 37 thousands, which are equal to 3 ten-thousands and 7 thousands.
Writing 7 as the thousands' figure of the amount, we add 3 ten-thoa
sands with the figure of the ten-thousands column; thus, 3, 11, 19, 22,
24, 31 ten-thousands, which are equal to 3 hundreds thousands and 1
ten -thousand ; and as there are no higher denominations, we write the 3
and 1 in their appropriate places.
Having thus added all the denominations, we must have the sum, or
amount of the numbers, which is 317,791.
Note. — Many call the names of the separate numbers added, as
well as the results of the addition, and would add, thus: 6 and 3
• If the learner does not readily understand this method of addition,
let him for a time call the separate numbers added, as explained in the note*
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44 ADDinON.
Are 9. and 5 are 14, and 8 are 22, and 4 are 26, and 7 are 33, and 8 an
41 ; 41 units ore equal, &<:. This method is much less expeditious than
the first one given, and therefore should not be adopted. Indeed, we
may say, that calling the names of the separate numbers comprising a
nam is to addition what spelling, i. e., calling the letters composing a
word, is to reading.
ffl. Compound Addition,
(a.) Compound Addition is the addition of compound
oumbers.
What is the sum of £8 15 3. 11 d. 2 qr. -j- £3 12 s. 8 d. 3 qr.
+ £9 19 s. 9 d. 1 qr. + £7 18 s. 10 d. 3 qr. + £5 18 s. 2 qr.
+ £6 8d. Iqr. + £5 138. 3d. + 16s. 8d. Iqr.?
Solution. — ^We first write the numbers, placing pounds under pounds,
■hillings under shillings, &c., in order that figures expressing the same
denomination may be near each other.
Jb
s.
d.
qr.
8
15
11
2
3
12
8
3
9
19
9
1
7
18
10
3
5
18
2
6
8
1
5
13
3
16
8
1
48 16 1 = Amount
Beginning with the right hand column, as before, and reducing as we
add, we proceed thus : 1 qr. and 1 qr. are 2 qr., and 2 qr. are 4 qr. va
1 d., and 3 qr. are 1 d. 3 qr., and 1 qr. are 1 d. 4 qr. = 2 d., and 3 qr. are
£ d. 3 qr., and 2 qr. are 2 d. 5 qr. = 3 d. 1 qr.
Writing 1 as the farthings' figure of the sum, we add the 3 d. with
tfio nnmf>ers in the pence column, thus: 3 d. and 8 d. are 11 d., and
5 d. are 14 d. = 1 s. 2 d.,* and 8 d. are 1 s. 10 d., and 10 d. are 1 s. 20 d.
•c 2 s. 8 d.,* and 9 d. are 2 s. 17 d. = 5" s. 5 d.,* and 8 d. are 3 s. 13 d.
'«s4 8.1 d.,* and 11 d. are 4 s. 12d.* = 5s.
Writing as the pence figure of the sum, we add the 5 s. with the
-^Vambers in the shillings column, thus: 5 s. and 16 s. are 21 8. = jC1
• Since 12 d. s 1 1
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ADDITION. 4v
1 s.,* and 13 s =£1 14 s., and 18 s. arc £1 32 8.* = j£2 12 8., and 18 8
are £2 30 s. = £3 10 s. * and 19 s. are £3 29 s. = £4 9 s^* and 12 8
are £4 21 s. = £5 1 s.,* and 15 s. are £5 16 8.
Writing 16 as the shillings' figure of the sum, we add £5 with tha
numbers of the pounds column, thus : 5, 10, 16, 21, 28, 37, 40, 48.
As all the denominations of the given numbers have been added, the
amount sought must be £48 16 8. d. 1 qr.
(h,) In the above form of solution, the numbers to be
added have been named merely to insure that the explana*
tions should be understood, but in practical work the addition
should be performed by naming only results.
Thus : 1 qr., 2 qr., 4 qr. = 1 d. ; 1 d. 3 qr., Id. 4 qr. = 2 d. ; 2 d.
3qr., 2d. 5qr =3d. 1 qr. AVritc 1 qr. 3 d., 11 d., 14 d. = l s. 2d.; I 8.
10 d., 1 s. 2) d. = 2 8. 8 d.; 2 s. 17 d. = 3 s. 5 d., &c.
59» Compound and Simple Addition compared.
(a,) Compound Addition involves precisely the same prin-
ciples that Simple Addition does. In both, numbers of the same
denomination are placed under each other, in order that they
may be more readily distinguished. In both, we commence
to add at the lowest denomination, in order to avoid the
necessity of erasing or altering figures which have been once
written ; in both, we reduce the sum of each column to units
of the next higher denomination, in order that the answer
may appear in its simplest form ; and in both we add the
units thus obtained with those written in the column of the
next higher denomination. Moreover, the same methods of
proof apply to both.
(b ) The slight difi'erences in the methods of applying these
principles result from the fact, that in simple numbers 10
units of one denomination always equal one of the next
higher, while in compound numbers there is no uniformity in
this respect.
Note to the Teacher. — The explanations and examples arc so
arranged that, should the teacher think it inexpedient ♦^ teach Com-
t)onnd Addition at the same time that Simple Addition is taught, he can
• Since 20 b. = £1.
Digitized by CjOOQ IC
46 ADDITION.
defer it till, in his opinion, the class are prepared for it We would^
however, recommend that whenever it is taught, it should be presented
as a further application of the principles involved in Simple Addinon.
The same thing may be said of Simple and Compound Subtraetion,
Multiplication, and Division.
S3* Methods of Proof,
(a.) Wi3 can test the correctness of the work in many
ways, a few of which we will mention.
First Method. — Go over the work carefully a second time
In the same manner as at first.
Second Method, — Begin to add at a different part of the
column from that at which the first addition was commenced
L e., if the first addition was commenced at the top, begin the
second at the bottom, and vice versa. This, by presenting
the figures in a different order, renders it improbable tliat any
mistake which may have been made in the first work will be
repeated.
Third Method. — Separate the numbers to be added ir.tc
two or more parts, add the parts separately, and then adJ
their sums. This method is illustrated below.
PHOOF OF EXAMPLE IN SO BY THIRD METHOD.
757981
24687
39764
B6328J
4395
283
86536 J
' 1st part.
2d part.
317791 = First Answer.
a = 226577 = Sum of first part.
b= 91214= « « second part
317791 = Sum of the two partial sums := first a "•-*w
Thus shewing that the work was correct.
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iU>DIT10N.
PROOF
OP
EXAMPLE IN Sil BY THIBD ME^IiOl^
€. 8.
d.
qr.
8 15
11
2
3 12
9 19
8
9
3
1
1st part
7 18
10
3.
5 18
2
5
5 13
8
3
1
^ 2d part.
16
8
1 .
48 16
1 =
= First Answer.
= Sum of 1st part
30 7
4
1 =
18 8
8
=
= Sum of 2d part.
47
48 16 1 = Sum of partial sums = first answer 6b«
tained. Thus showing the work to be correct
Note. — When long columns are to be added, it may sometimes b«
convenient to divide them in this way in performing the first addit.on.
The student should, however, accustom himself to adding the longest
columns without any separation into parts.
Fourth Method, — Beginning either at the right or at the
left hand to add, write the sum of each denomination scpa*
rately, and then add these sums together.
Fifth Method. — Begin with the left hand column, and pro*
ceed as follows : — >
PROOF OF THE EXAMPLE IN 50*
By adding the ten-thousands column we find that its sum is 28;
but as there are 31 ten-thonsands in the answer first obtained, we infer
that 3 ten-thousands were brought froni the lower denominations. 3
ten-thousands =^ 30 thousands, which, added to the 7 written in the
thousands* place of the answer, gives 37 thousands to be accounted for.
The sum of the thousands column is 34, which, taken from 37, leaves
8; thus showing, that if the work is correct, 3 thousands must have
been brought from the lower denominations. 3 thousands «=» SO him*
dreds, which, added to the 7 wnttep in the hundreds' place of the xmwet,
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48 ADDITION.
gives 37 hundreds to be accounted for. The sum of the hundreds
column is 33, which, taken from 37, leaves 4 ; thus showing, that if the
work is correct, 4 hundreds must have been brought from the lowe;
denomination. 4 hundreds = 40 tens, which, added to the 9 tens writ-
ten in the tens' place of the answer, gives 49 tens to be accounted for ^
The sum of the tens column is 45, which, taken from 49, leaves 4 ; thut
■bowing, that if the work is correct, 4 tens must have been brought
from the units column. 4 tens =: 40 units, which, added to the 1 unit
written in the units' place of the answer, gives 41 units to bo account-
edfor.
As the sum of the units column is 41, we infer that the work is
correct
PBOOF OF THE EXAMPLE IN 51*
By adding the pounds column, we find its sum is £43, which, taken
from the £48 written in the answer, leaves £5 ; thus showing, that if
the answer is correct, £3 must have been brought from the lower de-
nominations. £5 = 100 8., which, added to the 16 s. written in the
answer, gives 116 s. to be accounted for. The sum of the shillings
column is Ills., which, subtracted from the 116s., leaves 5s.; thus
showing, that if the answer is correct, 5 s. must have been brought from
the lower denominations. 5 s. = 60 d., which, as there are no pence
written in the answer, gives 60 d. to be accounted for.
The sum of the pence column is 57 d., which, taken from 60 d., leaves
8 d. ; thus showing, that if the answer is correct, 3 d. must have been
brought from the column of farthings j 3d. = 12qr., which, added to
the 1 qr. written in the answer, gives 13 qr. to be accounted for.
As the sum of the farthings column is 13, we infer that the answer is
correct
(h,) The first, second, and third methods of proof are the
most practical, but as the fourth and fifth furnish valuable illus-
trations of the nature of the various changes and reductions,
and call the reasoning faculties into healthful exercise, thej
should not be omitted by the student.
(c.) If, by any of these methods, we obtain a different result
from the one we first obtained, we may be sure there is an
error in one or both operations, and should examine both care-
fully to find it.
(d,) Some method of proof should always be resorted to,
until the pupil acquires sufiicient skill to be sure of the accu-
ra<7^ of his work without it.
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ADDITION. 49
t54. Importance of Accuracy and Certahtty,
(a.) No person who is willing to allow an error to pass undetebt-
CI can be a good arithmetician. Accuracy^ absolute accuracy, should
be aimed at in everj operation ; and no labor is too great which is
necessary to secure it. Not only shonld the results be accurate, ba^ the
tsmputer should know for himself that tJiey are so. If he has any dcubt
concerning a result, he should examine each and every step of hii wckI^
to see, .
First That it was a proper one to take.
Second. That it was taken at the right time.
Third. That it was taken correctly.
(b.) One problem thus solved and proved by a learner is of more real
value to him than ten solved by him and proved by another, or tested
by comparison with a printed answer. The accountant does not hesi-
tate to spend hours, and even days, in looking over long and complicated
accounts, to discover the cause of an error of a few cents in a trial
balance sheet,^ and surely the student ought not to shrink from the task
of proving the correctness of his solutions of the much more limple
problems contained in a school text book.
(c) Bapidity in the performance of numerical operations is scarcely
of secondary importance to accuracy and certainty. The most accurate
computers are usually the most rapid in their work.
* The trial balance sheet is used in keeping books by double entry, as
a means of determining whether any errors exist in the entries which have
been made in some given time, as a month, a quarter, (i. e., three months,)
six months, or a year. By its aid, the existence of an error may be ascer-
tained ; but the error itself cannot be discovered without examining the
separate entries and accounts.
An intelligent and highly accomplished accountant, who has charge of
the books of a large manufacturing establishment, employing three hundred
men, once spent nearly a week in examining his accounts, to discover the
cause of an error of a few cents ; and said he, " I never spent the same
amount of time more profitably." Another gentleman, bearing also a high
reputation, and receiving a good salary as an accountant, spent, to use his
own language, ** the greater part of four days in searching out the cause
of an error of ten cents." Both these gentlemen say, that if they should
adopt any other principle than that of absolute accuracy, they could not
retain their situations. Every accountant, business man, and practical
man bears similar testimony, and confirms these views. Indeed, most of
them say, that the knowledge of arithmetic acquired in the school room
has been of little practical value to them, because they did not learn to be
accurate and rapid in performing their work, and to know for themselvee
that they had been aoeurate.
5
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M
ADDITION.
SS» Problems for Solution.
Add the numbers in the following examples «—
!• 2. 3.
476392 832547 76486795
584273 482938 87630487
143253 548276 15327943
624415 398254 82753428
4.
5.
6.
206805795
6786.73
879.46
740376525
2974.37
82.375
814256324
8362.52
986.3
884567890
4593.67
69.486
825462813
6876.79
676.829
458269983
2394.16
403.586
7. .
8.
9.
4278.75
48.067
786.73
28.9878
2.6875
2.7598
6794.0000
138.2654
.58678
8a5946
63.4867
6.82694
S954.2765
867,9683
36.95006
6.7986
16.8787
.00487
^85
587.483
30.2857
69.8678
866.9486
6.7642
10.
11.
12.
8732.175
8.7037
49.0064
6248.262
2.9675
2.206
8711.547
86.08972
69.0425
8952.364
5.86102
87.63214
428.249
2.407967
88.8006
1497.168
82.27313
70.3728
6557.438
25.7529
74.85768J
4886.295
4.8063
428.4269
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ADDITION. fi]
13. ^Vlmt is the dum of 679487 + 886754 + 329687 +
435429 + 27G834 + 579487 ?
14. What is the sum of 824067 + 235143 + 543345 +
425341 + 876583 + 947869 ?
15 What is the sum of $478.87 + $526.94 + $857.93
+ $297.16 + $87.43 + $528.60 + $35.29 ?
16. What is the sum of 857436.57 + 25986.483 +
295463.867 + 297484.258 + 80672.005 ?
17. What is the sum of 258.647943 + 547.685329 +
27.84372 + 9765.4837 + 736.852066 + 542.063794?
18. What is the sum of 984137.612 + 257.00684 +
43687.5792 + 574869.23757 + 2068439.14238 + 1748.2
+ 13.37 ?
19. What is the sum of 1864 + 437.29 + 58.697 +
J2.86 + 7527.385 + 167.97 + 848.896 + 4.584?
20. ^Vhat is the sum of 389.40067 + 2768.4872 +
5894.276 + 1385.7281 ?
21. What is the sum of 728 + 436 + 549 + 278 + 867
+ 825?
22. What is the sum of 426764572681 + 894737629487
+ 179428630006 + 576428670639 + 584967245876?
23. What is the sum of 3798643 + 5978642 + 5489879
+ 675986 + 3768543 + 27864 + 3798742 + 8957387 +
9583796 + 8395989 + 3865372 ?
24. What is the sum of 83679 + 54873 + 72352 +
95873 + 8756 + 35906 + 87506 + 29764 + 38756 +
85742 ?
25. What is the sum of 57386 + 2864.8 + 879.86 +
28.697 + 5.4738 + .97986 + 7.5983 + 86.794 + 886.79
+ 2937.6 + 70003 + 9764.2 + 859.86 + 48.375 ?
26. What is the sum of $8.69 + $13.48 + $4.48 +
$8 64 + $37.15 + $47.13 + $.86 + $.25 + $9.37 +
$6.08 + $3.54 + $7.06 + $2.37 + $4.68 + $20.08 +
$7.57 + $7.48 ?
27. What is the sum of $4,175 + $3,867 + $5,384 +
$9,375 + $5.78 + $8,378 + $2,635 + $.875 + $1.25 +
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52 ADDITION.
$4.28 + $3.19 + $8,625 + $5,846 + $9,738 + $5.96 -|-
$7.50 4- $3.25 ?
28. What is the sum of $.27 + $.63 + $1.04 -f $.50 -f.
$.375 + $1.50 + $.07 + $.42 + $.625 + $.875 + $3.27
+ $5.94 + $.86 + $1.83 + $.06 + $.40 + $.125 -f-
$1.33 ?
29. What is the sum of $85.76 + $77.25 + $.86 -f.
$34.50 + $7.38 + $50.50 + $7.13 + $.47 + $-68 -f
$28.17 + $29.50 + $8.07 + $5.00 + $17.84 + $.03 H
$5.28 ?
30. What is the sum of 58694 + 67867.9432 + 45879.-
8376 + 28697.4 + 38679.58432 + 27598.542 + 36789.754
+ 58767.5437 + 86427.58697 + 98003.79 + 28547.3298
4- 28475.9767 ?
31. What is the sum of 958679.4437 + 298673.925 +
586732 + 9678.4593 + 486.7923 + 5878.6532 + 185.3
+ 28.6734 + 86.79635 + 28.76 + 59.836 + 45173.425 ?
32.
33.
£.
1. d.
qr.
£..
1. d.
qr.
17
13 11
1
164
13 8
1
18
15 8
3
231
6
3
29
19 6
1
485
19 11
2
47
8 10
3
738
18 2
3
25
13 11
2
487
16 10
87
18 9
3
833
19 11
3
34. 35.
T. cwt qr. lb. ob. T cwt qr. lb. oz. dr.
13 18 2 23 14 8 4 3 7 5 9
23 19 1 24 15 2 18 1 24 15 15
6 8 17 3 9 19 3 24 15 15
24 16 1 24 12 7 13 20 11 14
8 19 20 9 6 17 3 15 10 13
8 IS 3 22 13 7 14 1 22 11 12
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ADDITIOK
S$
86.
37.
lb. oz. dwt gr.
lb. oz. dwt gr.
4 6 18
23
5 9 19 21
2 11 13
5
2 11 13 23
1 7 8
21
1 8 6 12
8 16
20
2 5 9 6
1 10 9
17
3 10 19 19
5 9 18
20
8 7 17 22
88.
89.*
lb s S 9
««■•
m.
far. rd. yd. ft. in.
6 8 5 2
16
7
6 37 4 2 8
9 8 4 1
17
9
4 24 2 1 2
8 9 2 2
8
6
7 34 1 2 8
7 8 7 1
19
8
5 28 1 1 7
4 6 6 2
13
9
3 37 2 2 8
9 8 3 2
15
7
4 19 1 1 6
40.t
41.t
. ni. far. rd. yd.
ft. !n.
rd. yd. ft. in.
8 6 34 4
2 7
8 2 1 11
7 2 38 3
1 11
3 5 2 6
4 6 12 5
2 4
4 8 1 9
2 7 26 4
1 10
7 2 8
9 3 32 5
9
8 4 2 10
8 5 13 3
2 6
7 2 2 8
* In adding yards, it ynSl usually be well to consider every eleven yards
ts two rods. Such a course will, to a very great extent, avoid the use of
fractions. The pupil should, however, bear in mind that half of a yard a
1 ft. 6 in., and that when in any number there are 5 yards, and 1 ft. 6 in.
besides, the value may be better expressed as 1 rod.
t Obtaining the answer to the 40th example in the usual method, we
■hall find it to be 41 m. 7 far. 89 rd. /> yd. 2 ft. H in. This is correct, but it
is not in the best form, for although there are not units enough expressed
of any denomination to make one of the next higher, it equals 42 m. fur.
rd. yd. 1 ft. 5 in. Show the t^qt^ of this statement, and show also vfhy
the answer does not ctt Jlr^ appear in the best form.
J The awver to the 4l8t ej^amrte wiU take the form at first of 35 rd
4*
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64 ADDITION.
42. 43.*
m. fur. rd. yd. ft in. a. b. rd. yd. ft. in.
8 4 19 3 1 6 16 2 28 19 7 132
8 2 16 5 5 14 3 37 8 2 47
9 4 39 4 2 9 66 1 19 17 3 142
2 1 25 3 2 11 18 2 39 16 4 27
5 2 38 5 1 3 17 3 21 28 8 99
8 7 18 4 2 10 25 2 31 30 6 100
SO* Another, and often a shorter. Method of reducing
Compound Numbers.
It will often be more convenient to make the reductions by
adding enough of one number to another to give a sum equiv-
alent to a unit of the next higher denomination. We will
take, for illustratioQ, the example given in ff 1*
£.
1.
d.
qr.
8
15
11
2
S
12
8
8
9
19
9
1
7
18
10
3
5
18
2
6
8
1
5
13
3
16
8
1
48 16 1
Explanation. — Having found the sum of the farthings
eolumn to be equal to 3 d., we add the 3 d. with the column
of pence, thus : —
4i| yd. 3 ft. U in., which should be changed, for the sake of simplicity, to
35 rd. 5 yd. 1 ft. 6 in. Show the equality of the two expressions, and the
method by which the reduction can be made.
* In adding square yards, it will be of service to notice that 60| sq. yd.
K 2 sq. rd ; that 90} sq. yd. = 3 sq. rd. ; that 121 sq. yd. = 4 sq. rd ; and
thut I of a sq. yd. s 2 sq. ft, 96 s^. ^n* This will avoid an) ii^culty in thu
ns© of fractions. ^
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▲DDITIOK. 55
3 d. and 8 d. are lid., and 3 d. are 1 s. 2 d., (by adding 1 of the 3 d.
with the 11 d.,) and 8d. are 1 s. 10 d., and 10 d. are 2 s. 8 d., (by adding
3 of one 10 d. with tlie other 10 d.,) and 9 d. are 3 8. 5 d., (by adding 3 of
the 8 d. with the 9 d ,) and 8 d. are 4 8. I d^ (by adding 4 of the 5 d. with
the 8 d.,) and 1 1 d. are 5 s. d.
Adding the 5 s. with the shillings column, we have 5 s. and 16 s. are
£l 1 s., (by adding 4 of the 5 8. with 16 s.,) and 13 s. are £1 14 s., and
18 8. are £2 12 s., (by adding 2 of the 14 s. with 18 s.,) and 18 8. are
£3 108., (by adding 2 of the 12 s. with 18 5.,) and 19 s. are £4 9 s., (by
adding 1 of the 10 s. with the 19 s.,) and 12 s. arc £5 I s., (by adding 8
of the 9 s. with 12 s.,) and 15 s. are £5 16 8. = sum of shillings column.
(b,) We have mentioned the numbers added in order to
secure clearness of explanation, but in practical work the
results alone should be named.
Thus, beginning with the farthings, we have, —
1 qr^ 2 qr., 1 d., 1 d. 3 qr., 2 d., 2 d. 3 qr., 3 d. 1 qr. Write 1 qr.
3d., lid.. Is. 2d., Is. 10 d., 2s. 8d., 3s. Sd., 4s. id., 5 8. Od.
Write d.
5s., £1 Is., Id. 14s., £2 128., £3 lOs., £4 9 s., £5 1 8., £5 168.
Write 16 s.
The pounds are added as before.
ff 7. Common Method of adding Compound Numhers,
By the method of adding compound numbers which is com-
uonly given, the entire sum of each column is found before
educing to higher denominations. Tliis method, however,
will, as a general thing, be found much less expeditious than
either of the others.
It is illustrated in the following solution of the example
given in the last article.
Explanation. — By adding the farthings column, we find that its sura
is 13 qr., which, as 4 qr. = I d., must equal as many pence as there
are times 4 in 13, which are three times, with a remainder of 1. There-
fore, 13 qr. = 3 d. 1 qr.
Writing 1 as the farthings figure of the amount, we add the 3 d. with
tlie figures of the pence column ; this gives 60 d^ which, as 12 d. = I s.,
are equal to as many shillings as there are times 12 in 60, which are 5
times. Therefore, 60 d. = 5 s.
Writing as the pence figure of the amount, we add the 5 s. with the
figures of the shillings column. This gives 116 s., which, as 20 s. = £1
are equal to as many pounds as there are times 20 in 116, which are S
times, with a remainder of 16. Therefore, 116 s. = £5 16 i
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56 ADDITION.
Writing 16 8^ we add the £5 with the figures of the pounds colmniL
This gives jC48, which, being the highest denomination, we write.
As all the denominations have now been added, the sum or amount
must be £4S 16 s. 1 qr.
SS» Examples for Practice in the Methods of ffO^ ffl,
and tSO.
1. What is the sum of £4 17 8. lid. 2qr.-f £84 13s. 3d.
4- £7 19 s. 8d. 3qr. + il6 18 s. 9 d. 1 qr. + £7 15 8.1d,
+ £18 16 8. lid. 2 qr.?
2. What is the sum of £1386 15 s. 6d. + £3576 18 s. 10 d.
+ £463 19s. 4d. + £23 5 s. 8 d. + £648 48. 6d. +
£100 10 8. 3 d. ?
3. What is the sum of 40 lb. 7 oz. 5 dwt. 6 gr. -f 9 lb. 8 oz.
19dwt. 22 gr. + 21b. 11 oz. 19 dwt. 23 gr. + 7 lb, 8 dwt.
19 gr. + 11 oz. 6 dwt + 3 lb. 1 oz. 15 gr. + 8 lb. 17 dwt. +
8 lb. 23 gr. + 18 dwt. 7 gr. + 9 oz. 15 gr. -J- 7 lb. 3 oz.
13 dwt. 15gi\?
4. What is the sum of 18 w. 4 da. 21 h. 37 m. 5 sec +
37 w. 5 da. 16 h. 43 m. 57 sec + 19 w. 3 da. 14 h. 46 m.
38 sec + 19 w. 6 da. 23 h. 56 m. 27 see. + 43 w. 5 da. 2h.
17 m. 38 sec + 28 w. 1 da. 1 h. 5 m. 7 sec ?
5. What is the sum of 47 gal. 3 qt, 1 pt 2 gi. + 37 gal.
1 qt, 1 pt 1 gi. + 85 gal. 2 qt. 2 gi. + 25 gal. 2 qt. 1 pt. 3 gL
+ 54 gal. 2qt. 1 pt. 3 gi. + 18 gal. 2 qt. 1 pt. 2gi. + 37gal.
3qt. Opt. Igi. 4- 19 gal. 3qt. Ipt. 2gi. + 43 gal. Oqt.
6 pt, 3 gi. ?
6. What is the sum of 15 yd. 2 qr. 2 na. + 18 yd. 3 qr.
lna. + 27 yd. 3 qr. 3 na. + 42 yd. 1 qr. + 87 yd. 3na.+
3qr. 3na. + 26yd. 1 qr. 1 na. + 57yd. 3qr. 2 na. + 42yd.
Oqr. Ona. -f <54yd. 3qr. 3na.?
7. What is the sura of 181b 6 S 55 2 9 5 gr. + 7 tb 8S
75 19 18 gr. + 41b US 43 2 9 13 gr. + 25Ib 95
ID 4gr. + llg 19 + 21b 5§ 63 09 16gr. + 51b
lis 43 19 14gr.?
8. What is the sum of 2 m. 7 fur. 28 rd. 4 yd. 1 ft. 3 in. +
6m. 5 fur. 19 rd. 2 yd 2 ft. 11 in. + 25 m. 4 fur. 37 rd. 5 yd
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ADDITION. 57
8 in. + 94 m. 1 ftir. 24rd. 4yd. 2 ft. 8 in. + 14 m. 6 fur.
23rd. 2yd. Oft. 7 in. + 23m. 5 fur. 37 rd. 4yd. 1ft. lOin.
+ 57m.O fur. 33rd. 5yd. 1ft. lin.?
9. What is the sura of 19 m. 5 fur. 37 rd. 2 yd. 2 ft. 2 in.
■f IG m. 4 fur. 18 rd. 5 yd. 1 ft- 7 in. + 37 m. 15 rd. 2 yd.
2ft. 8in. + 17rd. 5yd. 7in. + 3m. 7fur. 18rd. 4yd.
2 ft. 9 v\. + 46 m. 3 fur. 13 rd. 2 yd. 1 ft. 9 in. -f 33 m. 4fur.
27 rd. 5yd. 1ft. 4in. + 19m. Ofur. 34 rd. 3 yd. ft;. lOin.?
10. What is the sum of 14 A. 3 R. 28 sq. rd. 27 sq. yd.
8 sq. ft. 12 sq. in. + 27 A. 2 R. 31 sq. rd. 17 sq. yd. 5 sq. ft.
137sq.in. + 35 A. IR. 3lsq. rd. 18 sq. yd. 5 sq.ft. 11 G
flq. in. -f- 21 A. 26 sq. rd. 25 sq. yd. 5 sq. ft^ 107 sq. in. -("
43 A. 2 R. 14 sq. rd. 19 sq. yd. -f 1 R. 15 sq. rd. 37 sq. in. ?
11. I bought some flour for $6.75 ; some cloth for $17.25 ;
a hat for $3.37 ; a coat for $19.42 ; a vest for $3.87 ; some
calico for $3.25 ; some flannel for $4.93 ; some silk for $23.99 ;
a pair of boots for $5.33 ; an overcoat for $22.75 ; a shawl
for $6.68 ; a pair of gloves for $1.46 ; an umbrella for $1.37;
and a pair of overshoes for $1.17. What was the amount of
my purchase ?
12. A trader sold 17 cases of broadcloth ; the first case
contained 317 yards, the second 296, the third 319, the fourth
339, the fifth 259, the sixth 347, the seventh 329, the eighth
286, the ninth 321, the tenth 294, the eleventh 337, the
twelfth 248, the thirteenth 324, the fourteenth 346, the fifteenth
299, the sixteenth 338, and the seventeenth 207. How many
yards were there in all ?
13. He received $984.36 for the first case, $849.23 for the
second, $1097.28 for the third, $1342.94 for the fourth,
$836.28 for the fifth, $1297.89 for the sixth, $1048.30 for
the seventh, $857.82 for the eighth, $1004.28 for the ninth,
$976.87 for the tenth, $1248.67 for the eleventh, $827.61 for
the twelfth, $1176.04 for the thirteenth, $1327.98 for the
fourteenth, $876.48 for the fifteenth, $1200.36 for the six-
teenth, and $758.93 for the seventeenth. How much did ho
jeceive for all ?
14. In the course of the year 1853, a flour dealer boughl
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08 ADDITION.
649 barrels of flour for $3798.75 ; 357 barrels for $2039.25 ;
439 barrels for $2679.00 ; 987 barrels for $6198.42 ; 299
barrels for $1925.37; 1168 barrels for $7385.94; 627 bar.
rels for $4369.27; 1359 barrels for $9967.84; 538 barrels
for $4279.63; 275 barrels for $2383.50; 96 barrels for
$816.00; 948 barrels for $8472.56; 358 barrels for $3615.80;
796 barrels for $8237.29 ; and 2962 barrels for $25,851.00.
How many barrels did he buy in all ? How many dollar^ did
lie pay for the whole ?
15. He gained $324.50 on the first lot ; $178.50 on the
second; $109.75 on the third; $740.25 on the fourth; $29.90
on the fiflh; $584 on the sixth; $600 on the seventh; $1359
on the eighth ; $470.75 on the ninth ; $277.75 on the tenth ;
$89.28 on the eleventh; nothing on the twelfth and thirteenth ;
$398 on the fourteenth; and $2154.25 on the fifteenth.
What was the amount of his ji^ains ?
S9» Addition of severed Columns at one Operation.
(a.) Accountants often add ti^v'O or three, and sometimes
four or more, columns of figures at a single operation.
(6 ) The fuUowing illustrates Rome of the methods of doing it : ^-
67
85
94
28
69
343
Explanation. — 69 plus 20 « 89, plos 8 =■ 97, pins 90 as 187, plus 4
i» 191, plus 80 =s 271, pins 5 = 276, plus 60 » 336, plus 7 » 343.
(c.) By adding tens first, and then units, as before, and naming onlj
results, we have 69, 89, 97, 187, 191, 271, 276, 336, 343.
(i.) A little practice will enable a person to add without separating
each number into tens and units, thus : 69, 97, 191, 276, 343.
(«.) After the student has become familiar with the
method of adding by single columns, he will find it a very
valuable exercise to add as above explained. We recommend
that he perform, at least, the first twenty examples under SS
by addirg two or more columns at a time.
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ADDITION. 59
G0« Leger Columns.
A great part of the work of an accountant consists in add-
ing long leger columns, like the following. Let the pupil find
the sum of the numbers in each, being as careful to obtain a
correct result as he would be if he were to receive or pay the
eeveral amounts.
1. 2- 8.
3.37
.7B
673.2B
A.33
•^7
597.B4
7.62
.63
3A26.B7
^B
2.75
2/9.4B
•97
/.20
B.37
2.50
A.37
/67.BA
6j9
B.29
69B6.32
/0.00
/3.B5
67A9.3/
A.23
2.00
AB63.27
■B.07
.62
7SA2.35
A.37
.25
29B6.2B
9AB
/.37
379.B7
A.2/
9.B3
2.59
/3.26
6.7s
69.BO
/^O
B.A3
AO6O.75
.57
20Ji.B
309.7^
3.0B
6.00
/2A.B7
A.96
/.oo
B520.06
:B5
/.50
2A93.2B
A.00
7-69
AB.75
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9
ADDITION.
4.
5.
6.
//A27.d/
2/7S.63
4/7B.3B
/3.B7
7AS.29
5/37.96
A^.OO
A.37
2000.00
67A.00
59.AB
/697.B/
2-75
S46.53
52B.63
^Ad3.S5
SS.47
542B.49
S7/^6
697.5S
954.B6
A.37
792.43
2797.7B
/^Ad.7A
/2A6.5B
934.67
6A7.S5
642.^7
52B.39
34.9B
42B.OO
776.95
s<^7.s6
/OOO.OO
B2.55
A^.S7
3B6.74
/67.73
34S.6A
^7.^9
4/27.4B
9-7^
4.26
29B.49
6.25
269.73
5B42.76
A327^69
^9-^7
37B.35
5/A.38
5B3.2B
49.27
693.27
679-^9
/B9.0/
A3.96
2B74.43
//0/.4B
279-Ba
B97.6/
69B.4/
67S6.39
2B54.55
64.B/
2BA.62
7443.75
5B7.6S
75.2B
4^.2B
/4.39
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SUBTRACTION. 61
SECTION V.
SUBTRACTION.
Gl* Definitions and Illustrations,
(a.) Subtraction is the process by which w«
riND THE DIFFERENCE OF TWO GIVEN NUMBERS, OR THA
EXCESS OF ONE GIVEN NUMBER OVER ANOTHER.
(b.) The following are questions in subtraction : —
Joseph had 34 apples, and gave awaj 6 of them. How manj did he
have left 1
Samuel had 16 cents and George had 9. How manj more had
Samuel than George ?
8 from 16 leaves how many?
How many are 12 — 81
(c.) The larger given number, or one from which we sub
tract, is called the Minuend; the smaller given number, or
one subtracted, is called the Subtrahend; and the result ob-
tained is called the Difference or Remainder.
Illustration. — In the first of the above examples 34 is the minuend, 6
is the subtrahend, and 28 is the difference or remainder.
(d.) The minuend and subtrahend must represent things of
the same kind, otherwise the subtraction cannot be performed.
Illustrations, — 5 apples from 7 apples leave 2 apples, and 5 pears
from 7 pears leave 2 pears ; but it would be impossible to take 5 pears
from 7 apples, or 5 apples from 7 pears. We cannot subtract 5 cents
from 7 dimes ; but if we should exchange one of the dimes for its value
in cents, we should have 6 dimes and 10 cents, from which if we should
inbtract 5 cents, there would be 6 dimes and 5 cents left.
We cannot subtract units from tens, but we can find how many units
% given number of tens is equal to, and then subtract from that number
of units.
63* Method of writing Nui/nbers and performing ProUemh
requiring no Reduction,
(a.) Although the result is not affected by the manner of
writing the numbers, it is convenient to place thos« of the same
6
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62 SUBTRACTION.
denomination near each other, or to write units under unite,
tens under tens, &c., in simple numbers, and pounds under
pounds, shillings under shill.ngs, yards under yards, feet under
feet, &c., in compound.
(b.) For the sake of uniformity, we usually place the
minuend above the subtrahend, and the remainder beneath;
beginning at the right hand to subtract.
(c.) The following examples will illustrate the application
of these principles : —
1. AVhat is the value of 4697 — 3265 ?
Solution. — Beginning at the right hand, and considering the denomi
nations separately, wo have 5 units from 7 units leave 2 units ; 6 tens
frora 9 tens leave 3 tens ;. 2 hundreds from 6 hundreds leave 4 hundreds;
8 thousands from 4 thousands leave 1 thousand.
Having thus subtracted the numbers in all the denominations, we
know that the remainder must bo 1 thousand, 4 hundred, 3 tens, and 2
units, or 1432.
The work would be written thus : —
4697 = Minuend.
3265 = Subtrahend.
1432 = Bemainder.
Second Example. — What is the value of £17 8 s. 9 d. —
£3 4s. 6d.?
Solution, — Beginning with the lowest denomination, we have 6d-
from9d. = 3d.; 4 8. from 8 8. = 4 s.; £3 from £17 = jei4.
The answer is, therefore, £14 4 s. 3 d.
Or, beginning at the left, £17 — £3 = £14; 8 8. — 4 8. =» 4 8.; •d.
— 6d.ss3d. ^ns. £14 48.3d.
The work would be written thus : —
£. 8. d.
17 8 9 Minuend.
3 4 6 Subtrahend.
14 4 3 Remainder.
63. Methods of Proof.
(o.) From the nature of subtraction, it is evident, that If .
the minuend were divided into two paits, such that oue
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SUBTRACTION. 68
flhonld equal the subtrahend, the other would equal the re-
mainder.
{b,) We have, therefore, the following methods of test-
ing the correctness of the work : —
First Method, — Add the remainder to the subtrahend, and
if the sum thus obtained is equal to the minuend, the work is
probably correct ; but if it is not, there is an error in either
the subtraction or the addition, and possibly in both.
Second Method. — Subtract the remainder from the minu-
end, and if the result thus obtained equab the subtrahend,
the work is probably correct.
WRITTEN WORK AND PROOF OF THE FIRST EXAMPLE.
4697 Minuend.
3265 Subtrahend.
1432 Remainder.
4697 Sum of Eem. and Sub. = Minuend.
3265 Diff. of Rem. and Min. = Subtrahend.
WRITTEN WORK AND PROOF OF SECOND EXAMPLE.
£, 8. d.
17 8 9 Minuend.
3 4 6 Subtrahend.
14 4 3 Remainder.
17 8 9 Sum of the Sub. and Rem. = Minuend.
3 4 6 Difference of Rem. and Min. = Subtrahend,
G4r* Problems requiring no deduction.
What is the value of each of the following ?
1. 854736 — 721423?
2. 9863764 — 420423 ?
3. 2948769 — 1432526?
4. $5476.92 — $1261.40?
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64 SUBTBACTION.
5. 736.87 yd. ~ 415.24 yd.?
6. 698.795 bu. — 243.521 bu. ?
7. 3 lb. 11 oz. 15 dwt 18 gr. — 1 lb. 8 oz. 13 dwt 4gr.?
8. 6T. 17cwt. 2qr. 261b. 13 oz. 11 dr. — 2 T. 6cwt
151b. 8 oz. 3 dr.?
9. 161b8§ 53 29 18gr. — 5lb4S 2 5 19 6gr.?
10. 1757 gal. 3 qt. 1 pt 3 gi. — 1323 gal. 1 qt 2 gi. ?
11. 18 rd. 4yd. 2ft. llin. — 6rd. 2yd. 1ft. 5in.?
SS» Simple Subtraction. Method when Reductions are
necessary,
Wlien, as is often the case, a figure in the subtrahend repre-
sents a greater value than the corresponding figure of the
minuend, we take one of a higher denomination in the minu-
end, reduce it to the required denomination, add its value to
the value of the figure already expressed, and subtract the
value of the subtrahend figure from the sum thus obtained.
First Example, — What is the difference between 62.7
and 35.86 ?
WRITTEN VTORK.
5 11 . 16,10 Minuend, changed in form.
6 2.7 Minuend,
3 5.86 Subtrahend,
2 6.84 Difference.
Explanation of Process. — As the!<3 are no hundredths expressed in
the minaend, we reduce one of the 7 tenths to hundredths, leaving 6
tenths. 1 tenth = 10 hundredths, from which subtracting 6 hundredths,
leaves a remainder of 4 hundredths.
As 8 tenths cannot be subtracted from 6 tenths, we reduce one of the
2 units to tenths, leaving 1 unit. 1 unit = 10 tenths, which added to
the 6 tenths left in the tenths' place equal 16 tenths ; 8 tenths from 16
tenths = 8 tenths.
As 5 units cannot be taken from the 1 unit left in the units' place,
we reduce one of the 6 tens to units, leaving 5 tens. 1 ten = 10 units,
which added to 1 unit equal 11 units j 5 units from 11 units = 6 unit»
8 tens from 5 tens leave 2 tens.
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SUBTRACTION. B9
The answer, then, is 2 tens, 6 units, 8 tenths, and 4 Dandredths, or
S6.S4.
This may be proved in the same waj that the preceding examplei
were.
Questions on the above.
Which expresses the larger number, 7 tenths, or 6 tenths and 10 hun-
dredths, and why? 2 units and 6 tenths, or 1 unit and 16 tenths ? 5
tens and 1 unit, or 4 tens and 11 units ? 6 tens, 2 units, and 7 tenths*
or 5 tens, 11 units, 16 tenths, and 10 hundredths ?
How then will the remainder, obtained by subtracting 35.86 from
62.7, compare with the remainder obtained by subtracting it from 5 tens,
11 units, 16 tenths, and 10 hundredths ?
Second Example, — How many are 83004 dollars minus
24765 dollars ?
WRITTEN WORK.
7 12 9 9 14 = Minuend, changed in form.
$8 3 4 z= Minuend.
$2 4 7 6 5 = Subtrahend.
$5 8 2 3 9 =r Remainder.
Erplatuxtion. — As we cannot take 5 dollars from 4 dollars, and as
there are no tens or hundreds expressed in the minuend, we take 1 thou-
sand from the 3 thousands, leavin^^ 2 thousands ; 1 thousand =» lO hun-
dreds, and taking 1 of these hundreds to reduce to tens, we have 9
hundreds left. 1 hundred = 10 tens, and taking 1 of these tens to
reduce to units we have 9 tens left. 1 ten = 10 units, which added to
the 4 units in the units' place = 14 units.
Now, by subtracting, we have
9 units from 14 units = 5 units.
6 tens from 9 tens = 3 tens.
7 hundreds from 9 hundreds = 2 hundreds.
4 thousands cannot be taken from 2 thousands, therefore we take 1
ten-thousand from the 8 ten-thousands in the minuend, leaving 7 ten-
thousands. 1 ten-thousand =10 thousands, which added to the 2
thousands in the thousands' place =12 thousands.
4 thousands from 12 thousands = 8 thousands.
2 ten-thousands from 7 ten-thousands = 5 ten-thousands.
The remainder is, therefore, 58239 dollars.
Questions upon the above. — How can it be shown that 3004 is eqaa]
to 2 thousands, 9 hundreds, 9 tens, and 14 units ? That 83004 is equal
to ten-thousands, 12 thousanas, 9 hundreds, 9 tens, and 14 units ?
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66 SUBTRACTION.
00* Compound Subtraction,
First Example, — Find the difference between 141b. 6o&
5 dwt 7 gr. and 8 lb. 7 oz. 18 dwt. 23 gr.
WRITTEN WORK.
13 17 34 41 Minuend changed in fonn.
lb. OS. dwt. gr.
14 6 15 17 Minuend.
8.7 18 23 Subtrahend.
5 10 16 18 Remainder.
Explanation, — As we cannot subtract 23 gr. from I7gr., we take
1 dwt from the 15 dwt, which reduced to grains and added to the 17gr.
equals 41 gr. ; 23 gr. from 41 gr. = 18 gr.
But as 18 dwt cannot be taken from the 14 dwt left in the minuend,
we take 1 oz. from the 6 oz. and reduce it to pennyweights ; 1 oz. at
80 dwt., which added to the 14 dwt equal 34 dwt \ 18 dwt from 34 dwt
■=6 dwt
As 7 oz. cannot be taken from the 5 oz. left in the minuend, we take
1 lb. from the 14 lb. and reduce it to ounces ; 1 lb. = 12 oz., which added
to the 5 oz. equal 17 oz. ; 7 oz. from 17 oz. = 10 oz.
8 lb. from 13 lb. = 5 lb.
The answer is, therefore, 5 lb. 10 oz. 6 dwt 8 gr., which maj be proved
as before.
Questions. — Which expresses the greater quantity. 15 dwt 17gr., or
14 dwt 41 gr., and why ? 6 oz. 14 dwt., or 5 oz. 34 dwt ? 14 lb. 5 oz.,
or 13 lb. 17 oz.? 14 lb. 6oz. 15 dwt 17 gr., or 13 lb. 17 oz. 34 dwt
41gr.1
How would the remainder obtained by subtracting 8 lb. 7 oz. 18 dwt
23 gr. from 14 lb. 6oz. 15 dwt 17 gr. compare with that obtained by
subtracting it from 13 lb. 17 oz. 34 dwt 41 gr. ?
Second Example, — A farmer took 8bu. 3 pk. 5qt of
com from a bin containing 17 bushels. How many bushels,
pecks, and quarts remained ?
Reasoning Process. — If the bin contained 1 8 bu., and he took oat
tbn. Spk. 5 qt, there would remain the difference between 17 bu and
8 bu. S pk. 5 qt This shows that 17 bu. is the minuerl, and 8 bu. 3 pk
f qt the subtrahend.
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SUBTRACTION. 67
16 3 8 Minuend, changed in form.
ba. pk. qt.
17 Minuend.
8 3 5 Subtrahend.
8 3 Remainder.
Explanation. — As there are no pecks or qaarts expressed in the
fflinaend, we take 1 bu. from the 17 bu. and reduce it to lower denomina-
tions. 1 ba. = 3 pk. 8 qt Therefore 17 bo. = 16 bo. 3 pk. 8 qt The
■abtraction can now be performed as before.
(a.) In examples involving fractional denominations, it will
UBually be more convenient to make all the reductions and
changes in the minuend before beginning to subtract, as in the
following : —
Third Example. — What is the difference between 8 id
Syd. 1ft. 4in. and2rd. 4yd. 2ft. 5in.?
"WRITTEN WORK.
7 8 2 10 = Minuend, changed in form,
rd. yd. ft. in.
8 3 14 = Minuend.
2 4 2 5 = Subtrahend,
.5 4 5 = Remainder.
Explanation, — Since there are more yards, feet, and inches expressed
in the subtrahend than in the minuend, we will take 1 rd. from the 8 rd.
and reduce it to lower denominations. 1 rd. = 5^ jd. == 5 yd. 1 ft. 6 in.*
which added to the 3 yd. 1 ft. 4 in. equal 8 yd. 2 ft 10 in. Therefore
8rd. 3yd. 1 ft 4in. — 2rd. 4yd. 2ft 5in. = 7rd. Syd. 2ft lOin.—
2 rd. 4 yd. 2 ft 5 in.
Note. — Had not the 1 rod been reduced to yards, and the i yard to
feet and inches, before commencing the subtraction, the answer would
have taken the form of ft rd. 3i yd. 1 ft 11 in., from which, by reducing
the i| yard to feet and inches, we should get 5 rd. 3 yd. 2 ft 17 in. sa 5 rd.
4 yd. ft. 5 in. = tlie answer obtained directly by first method.
67. Problems f<yr Solution,
What is the value of —
1. 3743 — 2554?
2. 839.74 — 213.78?
3. 37.9623 — 21.978 ?
4. 300.67 — 25.38?
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(58 SUBTRACTION.
9. 40006 — 14308?
10. 294.6 — 87.942 ?
11. 320.06 — 5.947?
12. 824.57 — 347.28?
5. .7285 — .0649?
6. .00632 — .00274?
7. 87432 — 52841?
8. 92846 — 24547?
What is the value of —
13. £17 68. 8d. Iqr.- £13 17 s. 3 d. 2 qr. ?
14. 25 cwt. 1 qr. 13 lb. 10 02. 7 dr. — 13 cwU 2 qr. 15 lb
80Z. lldr.?
15. 8 lb. 4 oz. 17 dwt. 13 gr. — 2 lb. 8 oz. 19 dwt. 20 gr. ?
16. 141b 9 S 45 2 9 12 gr. — 41b 10 3 43 2 9
16gr.?
17. 4147 bu. 1 pk. 4 qt — 2878 bu. 2 pk. 7 qt. 1 pt. ?
18. 49w. 3da. 19 h. 13m. 458ec. — 18w. 1 da. 22h.
40 m. 53 sec ?
19. £487 6 s. Od. 1 qr. — £236 11 s. 8d. 3 qr. ?
20. 27** 24' 47" — 19** 37' 51" ?
21. 38m. 4fur. 23 i-d. 4yd. Oft. 3in. — 32m. 5fiir
28rd.5yd. 1ft, 5in.?
22. 54m. 6ftir. 3rd. 8in. — 48m. 3 ftir. 8rd. 2yd.
2 ft;. 10 in. ?
23. 3 R. 14 sq. rd. 7 sq. yd. 2 sq. ft;. 19 sq. in. — 23 sq. rd.
11 sq. yd. 5 sq. ft. 138 sq. in. ?
24. 15 rd. 5 yd. 2 ft. 11 in. — 16 rd. 1 ft;. 4 in. ?
68. 77ie Cfhanged Minuend not usually written,
(a.) The changed form of the minuend has been written in
the preceding examples to insure that the nature of the reduc-
tions and changes shall be understood by the pupil. It is not,
however, customary to write it. The full explanation is the
Bame whether it is written or omitted ; but when it is omitted,
and eveiy stop of the process is understood and mastered, ab-
breviated explanations like the following may be adopted : —
WRITTEN WORK OP FIRST EXAMPLE UNDER 60*
62.7 = Minuend.
35.86 = Subtrahend.
26.84 = Remainder.
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SUBTRACTION. M
/fijf Abbreviated Explanation. — 6 hundredths cannot bo taken ftt>m
hundredths, but 1 tenth =10 hundredths, and 6 hundredths from 10
hundredths leave 4 hundredths.
8 tenths cannot be taken from 6 tenths, but one unit equal 10 tenths,
and 6 tenths added = 16 tenths. 8 tenths from 16 tenths = 8 tenths.
5 units cannot be taken from 1 unit, but 1 ten s= 10 units, ind I autt
added a=s n units. 5 units from 11 units leave 6 units.
3 tens from 5 tens leave 2 tens.
Hence the answer is 26.84.
Stcond Abbreviated Explanation. — 6 hundredths from 10 hundredths
leave 4 hundredths; 8 tenths from 16 tenths leave 8 tenths; 5 units from
11 units leave 6 units; 3 tens from 5 tens leave 2 tens.
This gives 26.84 for an answer, as before.
(i.) All explanations should finally be dropped, and only
results named, thus : — 4 hundredths^ 8 tenthsy 6 units j 2 tens *
giving for the answer 26.84, as before.
WRITTEN WORK OF FIRST EXAMPLE UNDER 66.
lb. oz. dwt. gr.
14 6 15 17 = Minuend.
8 7 18 23 = Subtrahend.
5 10 16 18 = Remainder.
First Abbreviated Explanation. — 23 gr. cannot be subtracted from
1 7 gr. ; but 1 dwt. s= 24 gr., and 1 7 gr. added are 41 gr. 23 gr. from 41 gr.
■"18gr.
18 dwt. cannot be taken from 14 dwt; but 1 oz. = 20 dwt., and
14 dwt added = 34 dwt 1 8 dwt from 34 dwt = 1 6 dwt
7 oz. cannot be taken from 5 oz. ; but 1 lb. = 12 oz., and 5 oz. added
are 17 oz. 7 oz. from 17 oz. = 10 oz.
Hence the remainder is 5 lb. 10 oz. 16 dwt 18 gr.
G9* Subtrahend Figure mag be increased instead of
diminishing Minuend,
It is obvious that the result would be the same, if, instead of consid-
ering the minuend figure of a denomination from which a reduction has
been made to be one less, we should consider the corresponding subtra-
hend figure to be one greater. In the former case, we subtract 1 (on
account of the reduced unit) before subtracting the subtrahend figure
while in the Wter we add 1 to the subtrahend figure, and subtract both
togetlier.
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70
SUBTRACTION.
Thus, in subtracting the tenths of the first example, we may subtract
1 tenth from the 7 tenths before subtracting the 8 tenths, or we may add
the 1 tenth to the 8 tenths, and subtract both at once. Many always
•ubtract by the last method. One method is as convenient as the other,
bat Ae one to which we are most accustomed will seem the easiest.
yO. Another^ and usually shorter^ Method of subtracting
Compound Numbers.
(a.) When in compound subtraction reductions are neces-
sary, and the changed minuend is not written, one of the
following methods is often, if not usually, easier than that
hitherto taken.
First Subtract from the value of the reduced unit, and add
the remainder to the minuend figure ; or,
Second. Subtract as many as possible from the written
minuend figure, and the rest from the value of the reduced
unit
(h,) Applying the first of the above methods to the example
just considered gives the following work : —
Subtracting 17 of the 23 gr. from the 17gr. leaves 6gr. to be taken
from 24 gr., (the value of the reduced unit.) 6 gr. from 24 gr. »> IB gr
Subtracting 14 of the ISdwt. from the 14dwt. leaves 4dwt. to be
subtracted from 20 dwt, (the value of the reduced unit) 4 dwt. from
20dwt = 16dwt.
Subtracting 5 of the 7 oz. from the 5 oz. leaves 2 oz. to be taken from
12 oz., (the value of the reduced unit.) 2 oz. from 12 oz. c= 10 oz.
8 lb. from 13 lb. leave 5 lb.
(c.) In practice, the above method may be abbreviated
thus : —
Subtracting 17 of the 23 gr. leaves 6gr., and 6gr. from 1 dwt, w
24gr., = logr.
Subtracting 14 of the 18 dwt. leaves 4 dwt., and 4 dwt from 1 oz., cir
2Cd^vt.,= 16 dwt
Subtracting 5 of the 7oz. leaves 2 oz., and 2 oz. from 1 lb., or 12oi.j
Ks 10 oz.
8 lb. from 13 lb. = 5 lb.
{d.) Abbreviating still more, we have —
17 from 23 = 6, and 6 gr. from 1 dwt, or 24 gr., = 18 gr.
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SUBTRACTION. 71
14 from 18 = 4, and 4 dwt. from 1 oz^ or 20 dwt , = 16 dwt.
5 fmm 7 ss 2, and 2 oz. from 1 Ib^ or 12 oz^ ss lo oz.
d lb. from 13 lb. =s 6 lb., giving, as before, 5 lb. 10 oz. 16 dwt 18 gr«
(f .) Adopting the second method, we have —
23 gr. from 1 dwt, or 24 gr., ss i gr^ which added to the 17 gr. glvM
ISgr.
18 dwt from 1 oz., or 20 dwt, =» 2 dwt., which added to the 14 dwt
gives 16 dwt
7oz. from 1 lb., or 12 oz., ss Soz., which added to the 5 oz. gives
10 oz.
8 lb. from 13 lb. <= 5 lb., giving, as before, 5 lb. 10 oz. 16 dwt 18 gr.
KoTB. — Practice will make the student so familiar with all these
methods, that he will be able to see at once which is best adapted to the
case ho is considering. ^
Tl, Problems for SohUion
What is the value of —
1. 3743 — 2917?
2. 94276 — 46324?
3. 867004 — 328527?
4. 8674.5 — 2594.326?
5. 45842.7 — 6243.984?
6. 7564.001 — 756.4002?,
7. 93.46 — 2.78457?
8. 6.0004 — 4.000563?
9. 100000000 — 7 ?
10. 7000000 — .000007 ?
11. 847.96 — 47.96823?
12. 487.6307 — 48.76307?
13. A land company bought 8479 acres of wild land, and
sold 3896 acres of it How many did they have left ?
Reasoning Process. — If they booght 8479 acres, and sold 3896 acres
of it, they must have left the difference between 8479 acres and 3896
acres, to find which we snbtract 3896 from 8479.
14. A 6h>p is valued at $27648, and its cargo at $49325.
How much more is the cargo worth than the ship ?
Reasoning Process. — If the ship is worth $27648, and th^ caigo is
worth $49325, the cargo must be worth as many dollars more than the
ship as there are in the difference between $49325 and $27648, to find
which the latter must be subtracted ^m the former.
15. Census returns show that the United States coritamed
3929827 inhabitants in the year 1790; 5305941 in 1800:
7239814 in 1810; 9638191 in 1820; 12866020 in 1830 j
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72 SUBTRACTION
17069453 in 1840; and 23263488 in 1850. How many
more did they contain in 1800 than in 1790 ?
16. How many more in 1810 than in 1800?
17. How many more in 1820 than in 1810 ?
18. How many more in 1830 than in 1820 ?
19. How many more in 1840 than in 1830 ?
20. How many more in 1850 than in 1840 ?
21. How many more in 1850 than in 1790 ?
22. How many more in 1850 than in 1800 ?
23. How many more in 1820 than in 1790 ?
What is the value of —
24. 27bu. 2pk. 2qt.— 18bu. 3pk. 3qt.?
25. 83 yd. 2 qr. 1 na. — 47 yd. 3 qr. 2 na. ?
26. 8cwt. 2qr. 151b. 7oz. 3dr. — 2cwt. 3qr. 241h.
Soz. 9dr.?
27. 37gaL 2 qt 1 pt 2gi. — 12 gaL 3 qt. 1 pt 3 gL
28. 871b.2dwt — 41b.7oz.5dwt 13gr.?
29. 24ft 3 S —5ft lis 45 23 Ugr.?
80. 83yd.— 2qr. 3na.?
81. £64 — £28 14s. 7d.2qr.?
32. 187 T. 3qr. 13 lb. — 67 T. 17cwt 1 qr. 181b. 6oz.
13 dr.?
33. 27 sq. rd. 5 sq. ft. I7sq. in. — 26sq. rd. 30sq. yd
7 sq. ft. 53 sq. in. ?
34- 6fur. 8rd. 3yd. 1ft. lin. — Ifur. 8rd. 4yd. 2ft.
11 in. ?
35. 18 yd. 1 na. — 14 yd. 2 qr. 3 na. ?
36. 6m. 1ft. — 5m.7fur.39rd.5yd. lft.2in.?
37. 18m. — 17m. 7fur. 39rd. 5yd. lft.5in.?
38. 231 A. 19 sq. rd. — 197 A. 3 R. 27 sq. rd. 15 sq. yd.
5 sq. ft. 97 sq. in. ?
39. 127** 18' 14" — 113** 47' 25"?
40. 307 T. 8 cwt 2 qr. 23 lb. 8oz. 12 dr. — 213 T. 15 cwt
231b. 11 oz. 6 dr.?
41. 527 yd. 1 qr. 2 na. 1 in. — 431 yd. 2 qr. 3 na. 2 in.?
42. 63m. — 27m.7fur.39rd.5yd.2ft.3in.?
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SUBTRACTION. 78
48. Bought 7 T. Idcwt 1 qr. 19 lb. of hay, from which T
iold 3 T. 7 cwt 3 qr. 26 lb. How much had I left ?
44. A goldsmith bought 7 lb. 7 oz. of gold. How much
will he have left after manufacturiug and selling 31b. 5gr«
of it?
45. A trader sold 9 yd. 3 qr. 2 na. of cloth from a piec9
containing 27 yd. Iqr. Ina. How much was left in tho
piece?
46. A man set on foot to travel from Boston to Springfield,
the distance being 98 miles. The first day he travelled 28 m.
7 fur. 19 rd., the second 24 m. 6 fur. 28 rd., the third 29 m.
4 fur. 36 rd. How far was he from Springfield at the end
of the third day?
47. A man undertook to dig a ditch for a certain price per
rod. On completing it he demanded payment for a ditch
87 rd. Oft;. 8 in. long. His employer, doubting his honesty,
measured it, and found it to be but 36 rd. 5 yd. 1 ft. 9 in. long.
A dispute arising between them, they called in Mr. Jenks to
settle it, agreeing to abide by his decision. He measured the
ditch, and found its length to be 36 rd. 16ft;.9in. What was
the difference in their measurements ?
7S8. Subtraction from Left to Right.
Wc can begin at the left to subtract as well as at the right, if we am
only careful to reserve one for redaction from each denomination in th%
minnend when it is required bj the lower denominations. This reduo-
tion will be necessary when the figures in the subtrahend at the right of
the denomination considered are greater than the corresponding ones of
the minuend.
Example, — How many are 508.935 — 249.748 %
508.935 as Minuend.
249.748 = Subtrahend.
259.187 as Remainder.
Explanation, — Keserving 1 hundred ftt>m the 5 hundreds, we naT«^
S hundreds fVom 4 hundreds =3 2 hundreds. Reducing the 1 hundred
Nserved to tens, and reserving 1 ten for further reduction, we have^ 4
7
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74 SUBTRACTION.
tens from 9 tens == 5 tens. Reducing the 1 ten to units, and addint^ 8
anits, we have, 9 units from 18 units = 9 units. Reserving 1 tenth, we
have 7 tenths from 8 tenths = 1 tenth. Reducing the 1 tenth rcserrod to
hundredths, and adding 2 hundredths, (1 hundredth being reserved, ) wo
have, 4 hundredths from 12 hundredths = 8 hundredths. Reducing I
hundredth to thousandths, and adding the 5 thousandths, we have, 8
thousandths from 15 thousandths = 7 thousandths. The answer is,
therefore, 259.187.
As soon as the process is sufficiently well understood to justify it, tho
explanations should be'omitted, and the results only named.
Thus, in the example explained above, the pupil should say, 2 hun-
dreds, 5 tens, 9 units, 1 tenth, 8 hundredths, 7 thousandths. The answer
is, therefore, 259,187.
It is also a good mental exercise to read the results at once by in-
specting minuend and subtrahend.
Having the numbers to be subtracted written thus, —
8274.01 == Minuend,
2357.23 = Subtrahend,
perform the subtractions mentally, beginning at the left, and read at
once 5916.78. Practice will make this very easy.
Perform the following subtractions by beginning at the left : —
1.
89704 — 29821.
6.
521.732 — 23.547.
a.
85.1 — 22.563.
7.
42.736 — 5.749.
3.
$56 — $8.73.
8.
$65.28 — $47.
4.
8006^5 — 32070.
9.
.4103 — .00627.
s.
472968 — 381489.
10.
678432 — 189146.
* 73. SvhtracHon of several Nurnbers at once.
A good method of proceeding when several numbers are to
be subtracted is, to subtract the sum of each column of the
subtrahend from the appropriate part of the minuend, redo*
cing and changing denominations, as before explained.
How many are 862 — 28 — 59 — 38 — 56 ?
WRITTEN WORK.
862 Minuend.
28^
59
\ Subtraherds.
56.
681 Remainder*
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SUBTRACTION. 75
Explanation. — Adding the units of the snbtrahcnds, we have, 6, 14,
S3, 31 units, which cannot be taken from 2 units. To obtain as many
as 31 nnits, wc.must take 3 tens from the 6 tens. 3 tens = 30 units,
which added to the 2 units equals 32 units ; 31 from 32 leaves 1.
Now, it makes no difference whether we take 3 tens (on account of
those we reduced to units) from the 6 tens, and afterwards take the tenf
01 the tens column of the subtrahend, or whether we take all together.
Aviopting the latter method, and adding the tens column, we have 3, 8,
11, 1G, 18 tens, which cannot be taken from 6 tens, but reducing 2 hun*
drcds to tens, and adding the 6 tens, we have 26 tens, from which 18
tens being taken, there will remain 8 tens.
2 hundreds from 8 hundreds = 6 hundreds. The answer is, there-
fore, 68'.
When the above is well understood, omit in practice a part
of the explanation, as follows : —
6, 14, 23, 31 from 32 leaves 1 unit ; 3, 8, 11, 16, 18 tens from 26 tent
IBS 6 tens ; 2 hundreds from 8 hundreds = 6 hundreds.
The following form may also be taken : —
6, 14, 23, 31, and 1 are 32 units ; write 1 in the units* place ;
3, 8, 11, 16, 18, and 8 are 26 tens ; write 8 in the tens' place;
2 and 6 are 8 hundreds ; write 6 in the hundreds* place.
Perform the operations indicated in the following examples
by the method explained above : —
1. 87642 — 273 — 4827 — 285 — 437 — 869 — 245.
2. 98402 — 2701 — 2596 — 1874 — 987 — 1283 —
5876.
3. 276.385 — 31.278 — 13.691 — 12.586 — 57.84 —
32.798 — 7.302.
4. 287000 — 328.7 — 221.3 — 344.37 — 2.851 —
317.06 — 576.823.
5. 283.587 — 1.27 — .328 — 9.063 — 57.063 — .00876
— 70.07 — .826.
6. Subtract .87 + 4.73 + 826 + 42.71 + 9.854 + 3.27
from 9012031.
7. Subtract 8837 + 1429 + 6372 + 8406 + 9785 +
4203 from 9120301.
8. Subtract. 8375.94 + 27f .483 + 5427.98 + 386.421 +
279.43 + 81 679 + .4237 -f- 4598.7 from 846271.3.
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76 MULTIPLICATION.
9. Mr. Ewell has in his possession $9478.63, but he owes
$143.27 to Mr. Webster, $549.71 to Mr. May, $581,375 to
Mr. Kmgsbury, and $378,875 to Mr. Bryant How much
will he have left after paying his debts ?
10. A man travelled 8725.67 miles in the following con-
Tcyances, viz. : 1285.89 miles in railroad cars, 876.81 miles
in a canal boat, 587.86 miles in a stage coach, 725.18 miles on
horseback, 647.25 miles on foot, 3147.82 miles in a steam-
boat, and the rest in a ship. How many miles did he travel
in a ship ?
11. Messrs. Howes and Baker bought 27147 bushels of
Indian com. After selling, at private sale, 1438 bushels to
one man, 2627 to another, 3781 to another, and 864 to an*
other, they sold the rest at auction. How many bushels did
they sell at auction ? They received $719 for the first lot,
$1313.50 for the second, $1890.50 for the third, $432 for the
fourth, and enough to make up $13573.50 for what they sold
at auction. How much did they receive for that which they
sold at auction ?
SECTION VL
MULTIPLICATION.
74:. Definitions and Illustrations,
(a.) Multiplication is a process by /vhich vtk
ASCERTAIN HOW MUCH ANT GIVEN NUMBER WILL AMOUNT
TO, IP TAKEN AS MANX TIMES AS THERE ARE UNITS IK
•OME OTHER GIVEN NUMBER.
(b,) The following are questions in multiplication : —
How many are 7 times 6 ?
What is the value of 9 multiplied by 6 1
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MULTIPLICATION. 77
What is the value of 7 X 5 ?
How mach will 8 books cost at 3 dolIai*s apiece !
(c.) The number supposed to be taken is called the mtiUi-
plicand^ tLe number showing how many times the multiplicand
is supposed to be taken is called the mtdtipliery and the result
is called the product.
(d.) The multiplier and multiplicand are called factors of
tlie product.
(e.) The product is said to be a mtdtiple of its factors.
Illustrations. — In the first of the above examples, 6 is the mnltipli*
eand, 7 is the multiplier, and the answer, 42. is the product 7 and 6 art
factors of 42, and 42 is a multiple of 7 and 6.
In the second example, 9 is the multiplicand, 6 is the multiplier, and
the answer, 54, is the product. 9 and 6 are factors of 54, and 54 is a
multiple of 9 and 6.
(yi) The last example would be solved thus : —
If 1 book costs 3 dollars, 8 books will cost 8 times 3 dollars, which
are 24 dollars.
Here 3 is the multiplicand, 8 is the multiplier, and 24 is the product.
8 and 3 are factors of 24, and 24 is a multiple of 3 and 8.
(^.) In performing the operation, the multiplier must
always be regarded as an abstract number.
Illustration. — A number can be taken 3 times, 5 times, or 8 times,
but it would be absurd to speak of taking it 3 bushels times, 5 houses
times, or 8 books times.
(h,) The product must be of the same denomination aa
the multiplicand.
Illustration. — 7 times 8 bushels = 56 bushels ; 9 times 4 books se.
36 books ; 7 times 5 tenths = 35 tenths, &c.
NoTB. — It must be observed, that there is an apparent exception to
the last statement, (A.) when the multiplier is a fraction, for j6 times .04
Ks X)24 ; .02 times .0003 = .000006, &e. This will be explained in the
•ection on fractions. (See page 173, Note.)
(i.) To examine the nature of the operation on the numbers, let us
•appose that a person ignorant of all numerical processes, except that
»f counting, should be called upon to so've the last question.
7 ♦
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78 MULTIPLICATION.
(/) If he had a quantity of dollars, he might lay 3 in one place 4
more in anotl^cr, 3 more in another, and so go on laying 3 in a pli.d6
till he should have 8 piles of 3 dollars each. Since the dollars in each
pile would bay 1 book, the dollars in all would buy 8 books ; he might
then, by counting the dollars in the 8 piles, find how much the books
would cost.
(k.) If ho should have no dollars, he might still determine the result
in a similar way, by using pebbles, sticks, marks, or any thing else of a
like character. After learning how to add, he might obtain the result
by adding 8 threes together.
(/.) If, after having obtained the result in some way similar to the
above, he should remember it, he would ever afker be able, without count-
ing or adding, to give the answer to any question requiring the amount
of 3 taken 8 times.
(m.) If he should learn in a similar manner the several amounts of
10 and each number below 10, taken as many times as there are units in
each successive number from 1 to 10, he would learn the common mul-
tiplication table as far as ten. If he should now learn how to apply
this knowledge to the decimal system of numbers, he would be master
of the process of multiplication.
KoTB. — The very common definition. "Multiplication is a short
method of addition," is not a good one, any more than would be, " Mul-
tiplication is a short method of counting ; " for while it is true that the
results obtained by multiplication might be obtained by addition, it is
equally true that they might be obtained by counting. It is true that
multiplication has a dependence both on addition and counting, but it is
equally true that it is as distinct from them as they are from each other,
and that when we multiply we neither add nor count.
For instance, when we find the sum of 75798 + 24687 -f- 39764 -f-
86328 -f 4395 + 283 + 86536, by the method explained in article 50, we
add them ; but when we merely remember, and state that their sum is
817791, we do not add them.
So when we call to mind that 4 and 4 are 8, and 4 are 12, and 4 are
16, and 4 are 20, we add 5 fours together ; but when we merely remem-
l>er that 5 fours, or 5 times 4, are 20, we perform no addition, although
as a result, we h^ve in the mind the sum of 5 fours.
VS. Pf^oduct not qfepted ^ Change in Order of Factwrh
(a.) In determining the product of two numbers, it iQQ.ko8
no difference which is regarded as the multiplipanfl, prpyi4pj
the other is regarded as the multiplier.
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MULTIPLICATION. 79
Thus : 6 times 4 = 4 times 6, or 6 fonrs = 4 sixes = 24.
Again : 5 times 3 = 3 times 5, or 5 threes = 3 fives = 15.
(6.) The principle may be proved true for all numbers, by the follow-
ing arrangement of dots : —
Considering the dots as being arranged in horizontal rows, there are
8 rows with 5 dots in each row ; considering them as being arranged in
Tenical rows, there are 5 rows with 3 dots in each row j and reckoning
in either way we include all the dots.
(c.) Now, if these rows were extended in either direction, always
being kept equal to each other, it is evident that the number of rows
reckoned in one direction would always be equal to the number of dots
that would be in a row were the rows reckoned in the other direction,
and that all the dots would be reckoned in both instances. The number
that represents the multiplicand when the rows are reckoned in one
direction, will represent the multiplier when they are reckoned in the
other, while the product, or number of dots, will be unaltered.
(d.) Hence, it must always be true that it makes no difference with the
product which of the two factors is taken for a multiplier, provided the
other be taken as the multiplicand. It will generally be most convenient
to consider the larger factor as the multiplicand, though not always so.
Note. — In changing the order of factors, the one taken for the
multiplier should always be regarded as an abstract number, (see 74,^,)
while the other should take the denomination of the original multipli-
cand. Thus, 4 limes 3 apples = 12 apples ; or changing the order of
the factors, we should have 3 times 4 apples = 12 apples. In the first
case, 4 is &n abstract, and 3 a concrete, number j but in the second, 4 is
a concrete, and 3 an abstract, number.
So 6 times $8 = 8 times $6 ; 4 times $.09 = 9 times $.04 ; 5 times
$4.6j = 469 times $.05 ; &c.
76. Simple MuUipUcation. — When only one Factor is a
large Number,
When either factor is a large number, it will be well to
consider its denominations separately, and, if we write the
resells as we obtain them, to begin with the lowest denomina*
tion.
What will 7 acres of land cost at $75.69 pei acre ?
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80 MULTIPLICATION.
Keaaoning Process. — If 1 acre costs $75.69, 7 acres will cost 7 timei
$75.69, which can be found by multiplying it by 7.
In performing the requisite multiplication, the nambers are usuallT
written in some convenient way, as the following : —
$75.69 = Multiplicand.
7 = Multiplier.
$529.83 = Product.
Explanation. — Beginning at the right hand, or lowest denomination
of the multiplicand, we have 7 times 9 hundredths = 63 hundredths, or
6 tenths and 3 hundredths.
Writing the 3 in the hundredths' place, and reserving the 6 tenths to
add to the product of the tenths by 7, we have 7 times 6 tenths = 42
tenths, and 6 tenths added, = 48 tentiis, = 4 units and 8 tenths.
Writing the 8 tenths, and reserving the 4 units to add to the product
of the units by 7, we have 7 times 5 units = 35 units, and 4 units added,
t= ^9 units, = 3 tens and 9 units.
Writing the 9 units, and reserving the 3 tens to add to the product
of the tens by 7, we have 7 times 7 tens = 49 tens, and 3 tens added, ■=
52 tens, which, being our last product, we write. The result, then, is
629.83.
Note. — As soon as practicable, the explanation should be abbre-
viated, so as to name only results. Thus, 63 hundredths ; 42, 48 tenths ,
85, 39 units ; 49, 52 tens. Ans. $529.83.
77. Multiplication of Compound Numbers,
What will 8 casks of wine cost at £3 9 s. 7 d. per cask ?
Reasoning Process. — If 1 cask costs £3 9 s. 7 d., 8 casks will cost 8
times £3 9 s. 7 d., which can be found by multiplying it by 8.
WRITTEN WORK.
£. S. d.
3 9 7 =r Multiplicand.
8 = Multiplier.
27 16 8 = Product.
Explanation. — Beginning with the lowest denomination, we have 8
times 7 d. = 56 d., which, since 12d. = 1 s., must equal as many shil-
lings as there are times 1 2 in 56, wliich are 4 times, with a remainder of
8. Hence, 56 d. = 4 s. 8 d.
Writing the 8 d and reserving the 4 s. to add to the shillings of th«
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MULTIPLICATION. 81
next prodact, we have 8 times 9 shillings = 72 shillings, and 4 shillings
from the former product added, are 76 shillings, which, since 20 8 =»
£1, must equal as many pounds as there are times 20 in 76, which are
8 times, with a remainder of 16. Hence, 76 s. = jE3 16 8.
Writing the 16 s., and reserving the £3 to add with the pounds of the
next product, we have 8 times £3 = £24, and £3 added, = £27, which,
being the last product, we write.
78. Methods of Proof.
First Method. — Go over the work a second time in the
same manner as before.
Second Method. — ^^ Consider the multiplicand as the multi-
plier, and see if this gives the same result as before.
The figures being in this way presented in a different order, we shall
not be liable to repeat any mistake we may have made in the first work.
Third Method. — Write out by itself the product of the
multiplication of each denomination, beginning either at the
left or right, and afterwards add these products together.
The sum should equal the former product
Below is the written work of the examples in 76 and "77, as proved
by beginning at the left, and writing each denomination of the product
separately.
Example 1. 75.69 X 7
490. = Product of 70 by 7.
35. = Product of 5 by 7.
4.2 = Product of .6 by 7.
.63 = Product of .09 by 7.
529.83 s= Product of 75.69 by 7 = former Product
Example 2.
£3 98. 7d. X8
£24 == £24 = Product of £3 by 8.
72 8. =£3 12 8. = Product of 9 s. by 8
56d. == 4 8. 8 d. = Product of 7 d. by 8.
£24 72 8. 56 d. = £27 16 s. 8d. = Product of £3 9 8. 7 d
by 8 = former Product
Fourth Method. — Another method of proof is, after hav*
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82 MULTIPLICATION.
ing written out the work as at first performed, to begin at the
left hand, thus : —
75.69
7
529.83
7 tiroes 7 tens arc 49 tens ; but as there are 52 tens in the product, 3
tens must have come from the product of the lower denominations. S
tens = 30 units, and adding to this the 9 units written in the unit!
place, we find there ought to be 39 units in the product. 7 times 5 units
are only 35 units ; hence, if the work be right, 4 units must have come
from the product of the lower denominations. 4 units = 40 tenths, and
adding to this the 8 tenths written in the tenths' place of the product,
we find that there ought to be 48 tenths in the product. 7 times 6 tentliG
are only 42 tenths ; hence, if the product be right, 6 tenths must have
come from the product of the hundredths. 6 tenths = 60 hundredths,
and adding to this the 3 hundredths written in the hundredths' place, we
find that there ought to be 63 Inmdredths in the product; and as 7 times
9 hundredths are 63 hundredths, we may infer that the work is correct.
£. s. d.
8 9 7
8
27 16 8
8 times £3 = £24. But as in the written product there are £27,
£3 must have come from the lower denominations. £3 = 60 s., and
adding to this the 16 s. written in the shillings of the product, we find
that there ought to be 76 shillings in the product. 8 times 9 s. = 72 s. j
therefore, if the work be right, 4 shillings must have come from the
lower denominations. 4 s. = 48 d., and adding to this the 8 d. already
written, we find there ought to be 56 d. in the product; and as 8 times
7 d. = 06 d., we infer that the work is correct.
If the computer should find by multiplying numbers in one method a
result different from that obtained by multiplying the same numbers in
Bome other method, he may be sure that there is an error in one opera-
tion or the other, and he should examine his work patiently till Jc findi
It. No person who is willing to allow an error to pass undetected can
be a good arithmetician. (See 54.)
yO. Problems, — Multiplier a single Figure,
Note. — This article includes reducl jn descending. (See Not*
tflor 36th example.)
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MULTII ICATION. 83
1. What is the product of 84687 X 4 ?
2. What is the product of .0078673 X 7 ?
3. What is the product of 237.904 X 8 ?
4. What is the product of 20078 X 9 ?
o. What is the product of .00978 X 6 ?
6. What is the product of 796.783 X 7 '
7. What is the product of .00978 X 6 ?
8. What is the product of 37842 X 8 ?
9. What is the product of .7948 X 8 ?
10. 1 pound Avoirdupois of distilled water coniainfl
27.7015 cubic inches. How many cubic inches will 8 pounds
contain?
Reasoning Process. — If 1 pound contains 27.7015 cubic inches, 8
pounds will contain 8 times 27.7015 cubic inches, which can be found by
multiplying it by 8.
11. What will 7 acres of land cost at $94,839 per acre ?
12. 1 pound Troy of distilled water contains 22.794377
cubic inches. How many cubic inches will 8 pounds contain?
13. How many cubic inches are there in 7 cubic feet?
14. How much will it cost to build 9 miles of railroad at
$19783.27 per mile?
15. How much will 3 farms cost at $3879.86 each ?
16. How many feet would a man walk in 6 days at the
rate of 56487 feet per day ?
17. How many miles would a ship sail in 9 weeks, if she
sails at the rate of 1198.47 miles per week ?
18. How many inches are there in 4 miles, there being
63360 inches in 1 mile ?
19. How many pounds are there in 5 loads of hay, each
weighing 2794 pounds ?
20. How many acres in 7 lots, each containing 24.74386
acres?
21. How many square rods in a road 754 rods long and 4
rods wide ?
Reasoning Process, — Since a space 1 rod long and 1 rod wide con-
tains I square rod, a space 754 rods long and 1 rod wide must contain
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84 MULTIPLICATION
754 fsqaare rods, and a space 754 rods long aiid 4 rods wide must con
tain 4 times 754 sqanre rods, which may be found by multiplying 754
by 4. (See 40, Note.)
22. How many square feet in a walk 796 feet long and 9
feet wide ?
23. How many square feet in a wall 437 feet long and 6
feet high ?
24 Mr. Haven's garden is 124 feet long and 97 feet wide,
and is enclosed by a tight board fence 5 feet high. How
many square feet of boards are there in the fence ?
Suggestion to the Student. — Draw a plan of the garden.
25. Mr. Haven laid out a gravel walk, 4 feet wide, within
the fence, and extending all around the garden. How many
square feet did it contain ?
Suggestion to the Student. — Draw a plan of the garden and walk.
26. What will 49.67 barrels of apples cost at $3 per
barrel?
Reasoning Process.-^ If 1 barrel costs 3 dollars, 49.67 barrels will
cost 49.67 times 3 dollars, which is equivalent to 3 times 49.67 dollars.
The work would be written thus : —
$49.67 = Multiplicand.
3 = Multiplier.
$149.01 = Product.
Another Reasoning Process. — It is evident that if each barrel shonld
eost a dollar, all would cost as many dollars as there are barrels bought,
which in this instance would be $49.67. But if, at $1 per bbl., they oofi
$49.67, at $3 per bbl., they would cost 3 times $49.67.
The work may be written thus : —
49.67 bbl. at $3 per bbl.
$ 49.67 = cost at $1 per bbl.
S times $49.67 = $149.01 = cost at $3 per bbl. = An$.
27. What will 3749 lb. of saleratus cost at .08 per lb. ?
28. What will 178.69 bbl. of flour cost at $7 per bbl. ?
29. What will 27.96 firkins of butter cost at $9 per firkin?
80. What will be the weight of 3794 cannon balls, each
ball weighing 8 lb ?
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MULTIPl IC ATION. 85
81. If a soldier eats 4 lb. of meat in a week, how many
pounds will 2896 soldiers eat in the same time ?
32. What will 4736 casks of raisins cost at $7 per cask ?
33. J£ a vessel sails uniformly at the rate of 9 miles per
hour, how far will she sail in 176 hours ?
84. How many bushels in 1487 barrels, if each barrel
holds 3 bushels ?
35. How much will 3479 window weights weigh, if each
weighs 6 pounds ?
36. 82 bu. 3 pk. 5 qt. 1 pt. = how many pints ?
Reasoning Process. — Since there are 4 pecks for every bushel, there
must be 4 times as many pecks as bushels, or, in this case, 4 times 82
pecks, which are 328 pecks, and adding 3 pecks to this gires 331 peeks
as the value of 82 bn. 3 pk.
Since there are 8 quarts for every peck, therd must be 8 times as
many quarts as there are pecks, or, in this case, 8 times 331 quarts,
which are, &c
Another Form, — Since 1 bushel = 4 pecks, 82 bushels must equal
82 times 4 pecks, equivalent to 4 times 82 pecks, which are 328 pecks,
and adding 3 pecks to this gives 331 pecks, os the value of 82 bu. 3 pk.
Since 1 peck = 8 quarts, 331 pecks must equal 331 times 8 quarts,
equivalent to 8 times 331 quarts, which are, &c
NoTB. — Questions like the above, requiring that the value of num-
bers of one denomination shall be expressed in terms of some lower
denomination, are called questions in Keduction Descending; but as
they do not differ at all from other questions in multiplication, they
require no separate treatment. In performing them, there is no need of
writing the multipliers, as they may be known from the table of the
weight or measure used. In reducing to any denomination of which
there are units already expressed, it will usually be more convenient to
Add those units at the time we make the multiplication.
METHOD OP WRITING THE WORK.
82 bu. 3 pk. 5 qt. 1 pt.
331 pk. = 82 bu. 3 pk.
2653 qt = 82 bu. 3 pk. 5 qt
5307 pt = 82 bu. 3 pk. 5 qt 1 pt
8
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S$ MULTIPLICATIOX.
The following abbreviated form may be adopted after th«
above is perfectly familiar : —
82 bu. 3 pk. 5 qt, 1 pt.
331 pk.
2653 qt
5307 pt = Ans.
87. 97 bu. 2 pk. 7 qt 1 pt := Jiow many pints ?
38. 187 bu. 1 pk. 6 qt pt. 3 gi. = how many gills ?
89. 238 gal. 3 qt 1 pt 2 gi. = how many gills ?
40. 596 gal. 2 qt 1 pt 3 gi. == how many gills ?
il. 9 lb 11 S '63 2 9 = how many scruples ?
42. 8!b3S73 1B = how many scruples ?
43. 937 le. 1 m. 5 fur. = how many furlongs ?
44 286 le. 2 m. 7 fur. = how many furlongs ?
45. What will 37 bu. 2 pk. 3 qt 1 pt of cherries cost at 4
cents per pint ?
46. What will 114 bu. 3 pk. 2 qt of wheat cost at 1 cent
per gill ?
47. What will 17 gal. 3 qt 1 pt 3 gi. of oil cost at 5 cents
per gill ?
48. What will 93 gal. 2 qt 1 pt 2 gi. of brandy cost at 8
cents per gill ?
49. What is the product of £22 18 s. 8 d. 2 qr. by 7 ?
£. 8. d. qr.
22 18 8 2 = Multiplicand.
7 = Multiplier.
160 10 11 2 = Product
Explanation. — 7 times 2 qr. = 14 qr., which (since 4 qr. =» 1 d.,)
equal as many pence as there are times 4 in 14, which are 3 times, with
a remainder of 2. Hence, 14 qr. = 3 d. 2 qr.
7 times 8 d. == 56 d., and 3 d. added == 59 d., which (since 12 d. sat
1 8.) equal as many shillings as there are times 12 d. in 59 d., which ara
4 times, with a remainder of 1 1. Hence, 59 d. = 4 8. 11 d.
7 times 18 s. = 126 s., and 4 s. added = 130 s., which (since 20 •
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MULTIPLICATION. 87
■B £\) eqaal us many pounds as there are times 20 in 130, which are <
times, witli a remainder of 10. Hence, 130 s. = £6 10 s.
7 times £22 = £154, and £6 added = £160.
Hence, the product is £160 10 s. II d. 2 qr.
Note. — When any denomination to be multiplied is very near a
unit of the next higher, the work may frequently be much shortened by
considering it a unit of that higher denomination, and subtracting for
its deficiency in value.
For instance, in the example above: since 18s. = £1 — 2s., 7
times 18 s. must equal £7 — 14 s., or £6 6 s., to which adding 4 s., (from
the previous product,) we have £6 10 s., as before.
What is the product —
50. Of £29 8 8. 11 d. 1 qr. multiplied by 9 ?
51. Of 37T. 19cwt 2qr. 241b. 11 oz. 7 dr. multiplied
by3?
52. Of 273 bu. 1 pk. 5 qt. 1 pt. multiplied by 2 ?
53. Of 9 lb. 8 oz. 13 dwt 22 gr. multipHed by 7 ?
54. Of 28 da. 17 h. 27 m. 58 sec. multipHed by 6 ?
55. Of 47 lb 8g 75 29 18 gr. multiplied by 4 ?
56. Of 238 gal. 1 qt. 1 pt 3 gi. multiplied by 9 ?
57. 0/ 674 lb. 4 oz. 19 dwt 20 gr. multiplied by 8 ?
58. Of 23 lb. 4 oz. 16 d>vt 22 gr. multiplied by 8 ?
59. Of 13 cwt 2 qr. 17 lb. 13 oz. 9 dr. multiplied by 6?
60. Of 6 T. 18 cwt 1 qr. 24 lb. 2 oz. 1 dr. multiplied by 7 ?
61. Of 9 bu. 3 pk. 7 qt multiplied by 9 ?
62. Of 9 gal. 2 qt 1 pt multiplied by 5 ?
63. Of 8 w. 1 da. 23 h. 59 m. 56 sec. multiplied by 7 ? *
64. Of £8 19 s. 11 d. 3 qr. multiplied by 8 ?
65. Of 9 T. 19 cwt. 3 qr. 24 lb. 14 oz. multiplied by 7 ?
66. Of 7 lb. 11 oz. 19 dwt 21 gr. multiplied by 4 ?
67. Of 483 yd. 3 qr. 1 na. multiplied by 9 ?
68. Of 4978 bu. 3 pk. 5 qt multiplied by 5 ?
69. Of 37 lb. 11 oz. 19 dwt 23 gr. multijiHed by 6 ?
70 Of £5871 18 s. 4 d. 1 qr. multiplied by 2 ?
• In performing this example, much labor may be saved by obsendng
that the multiplicand is only 4 seconds less than 8 w. 2da. Similar things
can frcquf ntly be applied %s in several of the subsequent examples.
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88
MULTIPLICATION.
80. Multiplication hy Factors.
(a.) It often happens that a number used as a multiplier is
the product of two or more factors. In such csises it is
sometimes convenient to resort to processes similar to those
explained below.
NoTB. — In writing the work here, as in several other places through*
out the book, we have used letters for convenience of indicating to the
eje the operations which have been performed, and the relations which
the numbers bear to each other. For instance, in the first four given
below, *' a = 743," means that the letter " a" stands for 743. " 2 X a =»
bs=s I486," means that two times the number "a," i. e., 2 times 743, =»
the number represented by "b," which is 1486. "6 X b = 12 X a,*
means that 6 times the number ** b " (i. e., 6 times 1486) equals 12 times
the number "a," (i. e., 12 times 743.)
The student will observe that the letter which in one form stands
for one number, may in another form stand for a different number.
Thus, in the first form "b" is used to represent 1486, while in the
•econd it is used to represent 2972.
How many are 12 times 743 ?
FIRST METHOD.
a= 743
2
SECOND METHOD.
a= 743
4
2 X a = b = 1486
6
4 X a = b = 2972
3
6Xb=12Xa = 8916 3Xb=12Xa = 8916
THIRD METHOD.
a= 743
3
3 X a = b = 2229
2
FOURTH METHOD.
a= 743
2 X a = b = 1486
2
arXb=6Xa=:c = 4458 2Xb=4Xa = c = 2972
2 3
2Xc=12Xa = 8916
X c= 12 X a =8916
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MULTIPLICATION. 89
Explanations,
First Method. — Since 12 = 6 times 2, 12 times a number most bo
equal to 6 times 2 times the number.
Second Method. — Since 12 = 3 times 4, 12 times a number must be
equal to 3 times 4 times the number.
ITiira Method. — Since 12 s= 2 times 2 times 3, 12 times a number
must be equal to 2 times 2 times 3 times the number.
Fourth Method. — Since 12 = 3 times 2 times 2, 12 times a number
must equal 3 times 2 times 2 times the number.
(b,) Solve the following examples in a similar manner.
What is the value —
1. Of 879 X 18?
2. Of 9874 X 27 ?
3. Of 8764 X 36?
4. Of 6427 X 42?
5. Of 4.379 X 64?
6. Of 2976.4 X 28 ?
7. Of 2377 T. 17 cwt 2 qr. 19 lb. 6 oz. 11 dr. X 63 ?
8. Of 27 lb 85 63 23 17 gr. X 45 ?
9. Of 19 w. 5 d. 17 h. 38 m. 29 sec. X 36 ?
10. Of £28 13 s. 10 d. 2 qr. X 35 ?
11. Of 48 lb. 10 oz. 16 dwt 19 gr. X 24 ?
12. Of 837 bu. 3 pk. 6 qt. 1 pt X 18 ?
(c.) The most useful application of the foregoing principle
18 made when the multiplier is some multiple of 10.
13. What is the product of 8746 X 400 ?
Solution. — Since 400 => 4 times 100, 400 times a number must equal
4 times 100 times the number: to obtain which we have only to remove
the point two places towards the right, (24) and multiply by 4. Hence,
8746
400
3498400
- 14. What is the product of 9.7487 X 7000 ?
Solution. — Since 7000 = 7 times 1000, 7000 times a number must
equal 7 times 1000 times the number ; to obtain which we have only to
multiply by 7, and remove the point three places towards the righti
Ileuce.
9.7487
7000
68240.9
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90 MUI.TirLICATION.
(d.) In like manner, to multiply by 60, wc may multiply
by 6, and remove the point one place towards the right ; to
multiply by 9000000, we may multiply by 9, and remove the
point six places towards the right. In any case, all places
lefl vacant between the number and the point must be filled
with zeros. (See IS.)
What is the product —
15. Of 874379 X 20 ?
16. Of 27.9863 X 5000?
17. Of 714.26 X 90000?
18. Of 62.794 X 40?
19. Of 627.34 X 80?
20. Of 9137.6 -X 30000?
21. Of 84273 X 60?
22. Of 7643 X 7000000 ?
81. When both Factors are large Numhers.
(a.) "We can find the product of two numbers by multiply-
ing one of them by the parts into which we choose to sepai'ate
the other, and then adding the products thus obtained together.
Illustration. — 8 times 7=6 times 7 + 2 times 7 = 3 times 7 + 5
times 7 = 7 times 7 + once 7 = 4 times 7+4 times 7 ^ 56.
(h.) This principle, and the one illustrated in article 80,
are ordinarily employed when the multiplier contains several
denominations.
Illustrations. — We usnally get 83 times a number, by adding together
80 times the number and 3 times the number.
Wc get 647 times a number by adding together 600 times the num-
ber, 40 times the number, and 7 times the number.
We get 8009 times a number by adding together 8000 times the num-
ber and 9 times the number.
(c.) It can make no difference which part of a number we
multiply by first, provided we multiply by all its parts ; yet
for the sake of uniformity it may be well generally to begin
with the lowest denomination.
1. Suppose that we are required to find the product of
5794 X 78?
Explanation. — We may find the product by 8 in the usual manner.
To find the product by 70, we have only to multiply by 7, and remove
the point one place towards the right, or, which is the same thing, the
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MULTIPLICATION. 91
tigaren one place towards the left Adding these results together will
give 78 times the number.
WBITTI.N WORK.
a = 5794 = Multiplicand.
78 = Multiplier.
8 X a = b = 46352 = Product by 8.
70 X a = c = 405580 = Product by 70.
b + c=78Xa = 451932 = Product by 78.
(d.) Since the product of the multiplication by the units is
sufficient to ^x the place of the figures in the subsequent
products, the zero at the right of the second product need nol
have been written. The product would then stand, —
4345 zz: Product by 5.
6083 = " « 70.
65175 z= ^' « 75.
(e.) Examples in which the multiplier consists of mora
than two figures are performed in a similar way.
2. What is the product of 780.69 X 20850 ?
WRITTEN WORK.
a = 780.69 z= Multiplicand.
20850 = Multiplier.
a X 50 = b = 39034.50 = Product by 50.
a X 800 = c = 624552. = Product by 800.
a X 20000 = d = 15GI38 = Product by 20000.
b + c + d = 16277386.50 = Product by 20850.
Note. — In prc.2tice, only the necessary figures should be writtea
TTrns; —
780.69
20850.
39034.50
624552.
156138
162773S6.50
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98 MULTIPLICATION.
What is the product of —
8. Of 80276 X 89 ?
4. Of 298794 X 148 ?
5. Of 273986 X 27?
6. Of 4943 X 78?
7. Of 23879 X 2741 ?
8. Of 84976 X 203?
9. Of 25873 X 506 ?
10. Of 47.296 X 37 ?
11. Of 423758 X 6200 ?
12. Of 594.27 X 80200 ?
13. Of 30678 X 427 ?
14. Of 5.796 X 238?
15. Of 45.718 X 482?
16. Of .00876 X 74?
17. Of 67.968 X 327 ?
18. Of 970062 X 37?
19. Of 29743 X 806?
20. Of 8427 X 3076 ?
21. Of 279437 X 27623 ?
22. Of 8.64298 X 43000 ?
(/.) It is obvious that the product obtained by multiplying
one number by the difference of two numbers, is equivalent
to the difference of the products obtained by multiplying the
numbers separately by the two numbers.
Illustratum. — 5 times 3 = 7 times 3 — 2 times 3 = 8 times 3 — 3
times 3 := 15 times 3 — 10 times 3 = 29 times 3 — 24 times 3, &<:.
(^.) This principle is the reverse of that stated in (a,) and
can often be advantageously applied, as illustrated below.
(A.) To multiply by 99, observe that since 99 = 100 — 1, 99 times a
number mast equal 100 timcs^ the number minus once the number. For
•xample, —
99 times 837 = 100 times 837 — 837 = 83700 — 837 = 82863.
(i.) To multiply by 999, observe that since 999 = 1000 — 1, 999
times a number must equal 1000 times the number minus once the num-
ber. For example, —
999 times 14.67 = 1000 times 14.67 — 14.67 = 14670 — 14.67 =a
14655.33.
(y.) To multiply by 699, observe that since 699 = 700 — 1, 699
times a number must equal 700 times the number — once the number
For example,
699 X 5784 = 700 X 5784 — 5784 = 4048800 — 5784 = 4018016
(k.) In like manner we should have —
49 X 785 = 50 X 785 — 785.
98 X 4697 = 100 X 469? — 2 X 4697.
79996 X 394845 = 80000 X 394845 — 4 X 394845.
What i^ the product —
23. Of 48673 X 29? | 24. Of 37848 X 99?
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MULTIPLICATION. 95
25. Of 69435 X 69 ?
26. Of 29485 X 999 ?
27. Of 7486 X 998 ?
28. Of 4278 X 3999 ?
29. Of 6786 X 49 ?
30. Of 4296 X 79 ?
31. Of 28643 X 999?
82. Ahhremated Method.
(a,) Wlien the multiplier consists of more than one de«
nomination, much labor in writing figures may be saved by
applying the principles illustrated in the following exam-
ples : —
1. What is the product of 8356 multiplied by 79 ?
Preliminary Explanation, — It is evident that, in perforroing the
required multiplication, we shall obtain units by multiplying 6 units by
9 units. We shall obtain tens by multiplying 5 tens by 9, and 6 units
by 7 tens, and we may have some from the product of the units. We
shall obtain hundreds by multiplying 3 hundreds by 9, and 5 tens by 7
tens, and wo may have some from the former products. We shall obtain
the other denominations in a similar manner. We may then proceed
thus, writing the figures of each denomination as usual : —
WRITTEN WORK.
8856 Multiplicand.
79 Multiplier.
660124 Product
Explanation. 9 times 6 units = 54 units. We write 4 units.
9 times 5 tens = 45 tens, -}- 5 tens (from the product of the units) =»
50 tens, -f- 7 tens times 6, or 42 tens, = 92 tens = 9 hundreds and 2
tens. We write 2 tens.
9 times 3 hundreds = 27 hundreds, + 9 hundreds (from the previous
product) == 36 hundreds, -|- 7 tens times 5 tens, or 35 hundreds, = 71
hundreds = 7 thousand and 1 hundred. We write 1 hundred.
9 times 8 thousands = 72 thousands, -f- 7 thousands (from the pre-
vious product) = 79 thousands, + 7 tens times 3 hundreds, or 21 thou*
sands, ==100 thousands = 10 ten-thousands. We write in the thou-
sands* place of the product.
7 tens times 8 thousands = 56 ten-thousands, + 10 ten-thousands
(from the previous product) = 66 ten-thousands, which wo write.
The multiplication being now completed shows the answer to b«
660124
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94 MULTIPLICATIOX.
(h.) The following exhibits the , necessary operations on
the numbers, and is practicallj a much more convenient solu-
tion than the full form above given.
^ 9 X 6 =s 54 units.
6 (from last prodact) + 9 X 5,+ 7 X 6 = 5 + 45+42 = 92 tens.
9 (from last product) + 9 X 3,H-7 X 6 = 9 + 27 +35 = 71 hun-
dreds.
7 (from last product) +9 X8,+ 7X 3 = 7 + 72 + 21 = 100
thousands.
10 (from last product) + 7 X 8 = 10 + 56 = 66 ten-thousands.
This gives for an answer 660124, as did the first method.
(c.) The last process being understood, the wo A may be
Btill further abbreviated by omitting to name the factors used.
Thus, 54 units = 5 tens and 4 unit?.
45 + 5 = 50, + 42 = 92. 92 tens = 9 hundreds and 2 tens.
27 + 9 = 36, + 35 = 71. 71 hundreds = 7 hundreds and 1 thou-
MDd.
72 + 7 = 79, + 21 = 100. 100 thousands = 10 ten-thousands and
thousands.
56 + 10 = 66. 66 ten-thousands.
Answer, as before, 660124.
((L) Finally, the work may be abbreviated so as to name
only results : —
54 units = 5 tens and 4 units.
45, 50, 92 tens = 9 hundreds and 2 tens.
27, 36, 71 hundreds = 7 thousands and 1 hundred.
72, 79, 100 thousands = 10 ten-thousands and thousands.
56, 66 ten-thousands.
Note. — The above methods are much more expeditious than is the
method of writing the product by each figure of the multiplier separate-
ly, and are no more liable to inaccuracy.
2. How much will 97 acres of land cost at $347 per acre?
8 If a cubic yard of sand weighs 2537 lb., how much will
fi8 cubic yards weigh ?
4. How many pounds are there in 18 T. 17 cwt. 1 qr. ?
5. If a ship sails 96 miles in one day, how far will she
ml in 247 day9 ?
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IIULTIPLICATION. D5
6. Bought 24 bundles of hay, each bundle containing 497 lb.
How many pounds were there in all ?
7. Bought 2947 gallons of oil at $.84 per gallon. How
much did it cost ? Sold it for $.97 per gallon. How much
was received for it ? What was the gain on it ?
8. Mr. Bussell bought 86 balls of twine, each ball contain-
ing 8794 ft., and Mr. Greene bought 57 times as much. How
many feet of twine did Mr. Kussell buy? How many did
Mr. Greene buy ?
9. How much will 83 casks of old wine cost at $138.47
per cask ?
10. How much will 67 tons of lead cost at $139.48 per
ton?
11. Mr. Hovey bought 6247 feet of land, and Mr. Ewell
bought 94 times as much. How many feet did Mr. Ewell
buy?
12. How many pounds are there in 958 boxes of sugar,
each box containing 743.67 lb. ?
Note. — The prodacts and sums employed in solving the aV>TQ
example are given below, but the pupil should be prepared to ip^« n
more full explanation.
56 hundredths = 5 tenths and 6 hundredths.
5 + 48 + 35 = 88. 88 tenths = 8 units and 8 tenths.
8 + 24 + 30 + 63 =s 125. 125 units = 12 tens and 5 units.
12 + 32 + 15 + 54 = 113. 113 tens = 11 hundreds and 3 tens.
11 4. 56 4- 20 + 27 == 114. 114 hundreds == 11 thousands and 4
hundreds.
11 + 35 + 36 s= 82. 82 thousands = 8 ten-thousands and 2 thou-
sands.
8 + 63 = 71. 71 ten-thousands.
The answer, therefore, is 712435.86 lb.
13. How many are 8795 times 96543 ?
Note. — By extending the principles before exjiUuned, we ,caii
the final product at once, as below.
96543
8795
849095685
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96 MULTIPLICATIOH.
bk the following fonns, two products are written :^
96543
8795
9171585 units
8399241 hundreds
« 95 X 96543.
= 8700 X 96543.
849095685
96543
8795
«B 8795 X 96543,
76751685 units
772344 thousands
= 795 X 96543.
=5 8000 X 96543.
849095685 &= 8795 X 96543.
14. If a cubic foot of iron weighs 486.25 lb., how much
will 847 cubic ft. weigh ?
15. How many square feet are there in a rectangular lot,
4827 feet long and 249 feet wide ?
16. How much will 48 acres of land cost at $28,968 per
acre?
17. What will 798 tons of hay cost at $14,278 per ton ?
18. I bought 287 bales of cloth, each bale containing
247.986 yards. How many yards did they all contain ?
19. How many square inches are there in a lot 247 ft.
longandl87ft. wide?
20. What will 47983 yards of doth cost at $2.83 per yd. ?
21. What will 7894 bbl. of flour cost at $6.37 per bbl. ?
22. How many solid inches in 5 C. 6 Cd. ft. 12 cu. ft. 1437
cu. in.?
28. How many dr. in 18 T. 16cwt. 1 qr. 141b. 6oz. lldr.?
24. A grain dealer sold 287 bushels of wheat at $1,294
per bushel, and 1479 bushels at $1,267 per bushel. What
did he receive for it ?
25. I bought 48 yards of broadcloth at $3,875 per yd.,
158 yards of doeskin at $1,166 per yd., and 379 yards of
cassimere at $1,125 per yd. What was the cost of the whole?
26. Mr. Aldrich owns 4 house lots, the first 328 ft. long
and 189 ft. wide ; the second 437 ft. long and 249 ft. wide ;
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BIYISION. 97
tbc third 129 ft. long and 88 ft. wide; and the fourth 97 ft;,
long and 86 ft. wide. How many square feet are there in all
of them ?
27. Mr. Whitney sold 84 acres of land at $34.96 per acre,
138 acres at $27.58 per acre, and 427 acres at $49.64 per
acre. How much did he sell the whole for ?
28. A city merchant went into the country to purchase
flour. He was absent from the city 27 days, and his expenses
while absent were $7,386 per day. He bought 175 bbl. of
flour at $5,875 per bbl., 516 bbl. at $5,948 per bbL, 1386 bbL
at $6.11 per bbl., and 827 bbl. at $6,087 per bbl. It cost
him $.634 per bbl. to get the flour transported to the city.
He sold 697 bbl. of it at $7,114 per bbL, 824 bbl. at $7,213
per bbl., and the rest at $6,978 per bbl. Did he gain or lose
by the adventure, and how much ?
29. A merchant bought 49.5 cases of cassimere, each case
containing 297 yd., at $1.1875 per yd. It cost him $.125 per
case to have the cloth removed to his store, and $.045 per
case to have it hoisted into his loft^ One case of the cloth
was stolen from him ; he sold 23 cases at $1,423 per yd., and
the remainder at $1,357 per yd., agreeing to deliver it at a
railroad depot, 1 mile ifrom his store. It cost him $.158 per
case to have it carried to the depot. Did he gain or lose on
the cloth, and how mu(^ ?
SECTION VII.
DIVISION.
Definitions and Ittiistrations,
(a.) Division is a process by which we ascertain
THE NUMBER OF PARTS OF A GIVEN SIZE INTO WHICH A
(»IV£N NUMBER MAT BE SEPARATED, OR BY WHICH WE
9
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98 Divisiow.
ASCERTAIK THE NUMB/.R OP UNITS THERE WILL B% W
KACH PART OBTAINED BY DIVIDING A GIVEN NCMilKB
JKTO A GIVEN NUMBER OP EQUAL PARTS.
(b.) The following are questions in division : —
1. 42 ^ how many times 6 1
S. What is the value of 54 dirided by 91
9. What is the ralue of 35 -!- 5 ?
4. How many apples at 3 cents apiece can be bonght for 24 cents *
5. What is J. of 36 ?
6. If 8 apples cost 24 cents, how much will 1 apple cost ?
(c) In the first of the above questions, we are required to find how
many 6's, or parts of 6 each, there are in 42 ; in the second, how many
9*8, or parts of 9 each, there are in 54 ; in the third, how many 5*s, or
parts of 5 each, there are in 35 ; in the fourth, how many 3*8, or parts
of 3 each, there are in 24 ; while in the fifth we arc required to find how
many units there will be in each part obtained by dividing 36 into 4
equal parts ; and in the sixth, how many units there will be in each part
obtained by dividing 24 into 8 equal parts.
{d.) This shows that there are two classes of questions in division,
viz. : one class in which, knowing the number to bo divided, and the num
her of units which each part is to contain, we are required to find the
number of parts j and one in which, knowing the number to be divided,
and the number of parts into which it is to be divided, we are required
to find how many units there will be in each part ; i. e., we are required
to find a fractional part of a number. Both classes of questions may
be solved by the same numerical process, though in their solution they
require different reasoning processes.
(e,) The number to be divided is called the dividend. In
the first class of questions the number which shows the size
of each part, and in the second class that which shows the
number of parts, is called the divisor. The result is called
the quotient,
Itltutrations, — In the first of the above questions, 42 is the dividend^ ,
• Is the divisor, and the answer, 7, is the quotient.
In the second, 54 is the dividend, 9 is the divisor, and the answer, 6,
li the quotient
In the fifth, 36 is the dividend, 4 is the divisor, and the answer, 9, is
OM quotient.
(/) The fourth example would be solved thus : —
Jf for 3 cents one apple can be bought, for 24 cents as many applei
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DIVISION. 99
can he bonpht as there are times 3 cents in 24 cents, which are 8 times.
Therefore, 8 apples can be bought for 24 cents, when 1 apple can be
bou^'ht for 3 cents. Here, 24 is the dividend, 3 is the divisor, and tho
answer, 8, is the quotient. The divisor and dividend are of the same
denomination, and the quotient is the number of times the divisor if
contained in the dividend, or the number of parts equal to the divisor
which the dividend equals.
(g,) The sixth example would be solved thus : —
If 8 apples cost 24 cents, 1 apple will cost 1 of 24 cents, which is 8
cci ts. Hence, I apple will cost 3 cents, if 8 cost 24 cents.
Here, 24 is the dividend, 8 is the divisor, and the answer, 3, is the
quotient. The dividend and quotient are of the same denomination,
and the divisor shows the number of times the quotient is taken in the
dividend, or the number of parts equal to the quotient which the divi-
dend equals.
(A.) To examine more fully the nature of the processes, let us see by
what methods the answers to the fourth and sixth questions could be
obtained.
(i.) It is ohyions that, to determine tho answer to the fourth ques-
tion, we must find how many parts of 3 cents each there are in 24 cents;
for each such part is the price of 1 apple.
We can do this by counting 24 cents into piles of 3 cents each, and
then counting the number of piles ; — by finding how many threes must
be added to make 24 ; — or by finding how many times 3 equal 24, by
our knowledge of multiplication. This last process is division.
(J.) To determine the answer to the sixth question, we must find
how many cents there will be in each part obtained by separating 24
cents into 8 equal parts.
We can do this by laying put 24 cents into 8 equal piles, and then
counting the number of cents in each pile : — or by finding, by our
knowledge of multiplication, what number must be taken 8 times to
equal 24; or, which will give the same numerical answer, by finding
how many times 8 equal 24. The last process is division.
NoTB. — It will be seen, that in the first class of questions, the divi-
•or and diWdend are of the same denomination, and that the quotient
expresses tho number of times the divisor is contained in the dividmidi
cr the number of parts equal to the divisor which must be taken to pn>-
dncc the dividend ; while in the second class, the divisor expresses th9
number of parts into which the dividend is to be divided, and the
quotient expresses the number of units in each part; thus making the
dividend ani quotient of the same denomination.
(k.) Practically, division is the reverse of multiplication. In th«
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100 ©ivisiox.
latter, the factors are given, and we are required to find the product
while in the former, one of the factors and their product are given, ana
we are required to find the other ; or, when the dividend will not exactly
contain the divisor, one factor and the product of the two pluH the
remainder are given, and we are required to find the other factor and
the remainder.
(/.) The divisor is always the given factor, the quotient is the re-
quired factor, and the dividend is the product, or the product plus the
remainder. The remainder is always less than the divisor.
(w.) The following examples illustrate this : —
1. "3 times 8," is a question in multiplication. The factors 3 and
8 ere given, and the product, 24, is required.
2. " How many times 8 = 24," or " 24 -5- 8 '" are questions in
division, in which the foctor 8, and the product 24, are given, and the
missing factor, 3, is required.
3. " ^ of 24, or 3 times what number = 24 ? " are questions in
division, in which the factor 3, and the product 24, are given, and the
missing factor is required.
4. " In 8 times 7, plus 5," the factors 8 and 7 are given, and their
product, plus 5, which is 61, is required.
5. " In 61 -f- 7," or " 61 =s how many times 7," the factor 7, and
the sum of the product, and remainder, which is 61, are given, and the
other factor and remainder are required.
6. " In ± of 61," or " 7 times what number = 61," the factor 7, and
the sum of the product, plus the remainder, are given, and the other fac-
tor and remainder are required.
84. Methods of Proof.
From the preceding illustrations it is evident, —
First. That, where there is no remainder, the divisor mul-
tiplied by the quotient must produce the dividend ; and that
the dividend divided by the quotient must produce the divisor.
Second. That if, when there is a remainder, the divisor and
quotient be multiplied together, and the remainder be added
to their product, the result will equal the dividend.
Tliird. That if the remainder be subtracted from the divi-
dend, and the remainder thus obtained be divided by tho
divisor, the result will equal the quotient; or, if it be divided
by the quotient, the result will equal the divisor.
Note. — The remain lei should always be less than the divisor; foi|
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DIVISION. 101
If it 13 not, the dividend will contain the divisor more times than is indl*
cated by the quotient figure.
8t5. Examples. — Quotient a single Figure.
(a.) How many oranges at 7 cents apiece can be bought
fxiV 61 cents ?
Reasoning Process. — If for 7 cents 1 orange can oo boaght, as many
oranges can be bought for 61 cents as there ai*e times 7 in 61.
Explanation. — To perform the necessary division, we observe that u
56 = 8 times 7, 61 must equal 8 times 7, with a remainder of 5; or,
since 5 -^ 7 = f , 61 == 8f times 7*
Hence, the quotient is 8, and the remainder is 7, or the completa
quotient is 8A, which shows that 8 oranges can be bought, leaving 5
cents unused, or that 8^ oranges can be bought with all the money.
First Proof, — See if the price of 8 oranges at 7 cents apiece, added
to 5 cents, will make 61 cents.
Second Proof. — Considering that 8i oranges are bought, see if they
^ill cost 61 cents at 7 cents apiece.
(6.) The above are proofs of the correctness of the entire work ; the
following only test the correctness of the division.
ITiird Proof. 8 times 7 = 56, and 5 added = 61 = dividend.
Fourth Proof 8 times 7 = 56 ; and 56 from 61 leaves 5 « re*
mainder.
Fifh Proof 5 from 61 = 56, and 56 -}- 7 = 8 = quotient
(c.) The work may be written thus : —
Dividend.
Divisor 7 ) 61 — 5 == Remainder.
8 = Quotient
(d.) Or, by expressing the division of the remainder, by
placing it in the form of a fraction, we havej —
Divisor 7 ) 61 = Dividend.
8f = Quotient
In the first form, the remainder is undivided, and is of the same de-
nomination as the dividend. It is placed opposite the dividend, with
the minus sign between, to indicate that all the dividend except tha$
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102
DIVISION.
has been divided. Indeed, it might hare been subtracted from the divi«
dcnd without affecting the quotient
In second form, the entire dividend is divided. Hence, the ^ is not a
remainder, but is a port of the quotient.
Note. — The distinction between a remainder and a fractional quo-
tient should be carefully observed.
What is the quotient —
1.
Of 37-
r4?
7. Of 43 ^ 7 ?
2.
Of 83-i
-10?
8. Of 26 -r- 3?
3.
Of 39-^
-7?
9. Of 47 -T- 5?
4.
Of 17-
1-2?
10. Of 48 -T- 9?
S.
Of 86-
1-9?
11. Of 75-^8?
6.
Of 57-
1-6?
12. Of 41 -^7?
13. How many apples at 3 cents apiece can be bought for
28 cents ?
14. How many barrels of flour at $9 per barrel can be
bought for $62 ?
15. How many yards of broadcloth at $6 per yard can be
bought for $58 ?
NoTB. — The student should practise on examples like the above,
till he can perform them with ease and rapidity.
86. Dividend a large Number.
(a.) When large numbers are to be divided, we begin with
the highest denominations, and proceed as in the following
example.
1. How many barrels of flour at $8 per barrel can be
bought for $58455 ?
Reasoning Process. — If for $8, 1 barrel can be bought, as many bar
l«ls can be bought for $58455 as there are times 8 in 58455.
WRITTEN WORK.
Dividend.
Divisor = 8 ) 58455 — 7 = Reminder.
7306 = Quotie?it.
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DIVISION. 108
Explanation of Process of Dividing, 8 is contained in 58 thoasandt
7 thonsand times, with a remainder of 2 thousands, (for 7000 times 3 =•
66000, and 2000 added = 58000.) We therefore write 7 as the thou-
sands' figure of the quotient
Re.ducinc: th« 2 thousands remaining to hundreds, and adding to
them the 4 hundreds, gives 24 hundreds to be next divided. 8 is con-
tained in 24 hundreds 3 hundred times, (for 300 times 8 ss 2400.) We
therefore write 8 as the hundreds' figure of the quotient.
As 8 is not contained as many as ten times in 5 tens, we write as
the tens' figure of the quotient, and reduce the 5 tens to units, consider-
ing them with the 5 units.
8 is contained in 55 units 6 times, with a remainder of 7 nnits, (for 6
times 8 = 48, and 7 added = 55.) We therefore write 6 as the units'
figure of the quotient
This gives 7306 for the quotient and 7 for the remainder ; or, since
7 -«- 8 = j^ it gives 7306J for a quotient
Hence, 7306 barrels of flour can be bought, leaving $7 unused ; or
7306 j. barrels can be bought with all the money.
The methods of proof are the same as those before given.
(J.) When the above explanation and the principles in-
volved are fully understood, the names of the denominations
may be omitted in writing the work. Thus : —
8 is contained 7 times in 58, with 2 remainder.
8 is contained 3 times in 24, with no remainder.
8 is contained times in 5, with 5 remainder.
8 is contained 6 times in 55, with 7 remainder.
Hence, the quotient is 7306, and the remainder is 7 ; or the complete
quotient is 7306Z..
(c.) In the following form, only the figures of the quotient
arc named.
7 thousands, 3 hundreds, tens, 6 units, and 7 remaining; giving
same result as before.
2. If 8 acres of land cost $9707, what will 1 acre cost ?
Reasoning Process. — If 8 acres of land cost $9707, 1 acre will cof t
I of «S707.
This may be found by dividing by 8, as in the last example ; or it
may bo found by the following method.*
* The student should observe that the real work performed, and th*
flgmea used in writing it, are the same by one method as by the other.
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104 DIVISION.
WRITTEN WORK.
Divisor 8 ) $9707 Dividend = cost of 8 acres.
$i213f Quotient = cost of 1 acre.
Wo divide thas : i. of 9 thousands = I thousand, with a remaindei
of I thousand. 1 thousand = 10 hundreds, and 7 hundreds added s= 17
hundreds. 4 of 17 hundreds = 2 hundreds, with a remainder of 1 hmi*
drcd. 1 hundred = 10 tens, and there are no tens to add. .| of 10
(ens = 1 ten, with a remainder of 2 tens. 2 tens = 20 units, and 7
nnits added = 27 units. X of 27 units = 3 units, with a remainder of
8 units.
Therefore, the quotient is $1213, and there is a remainder of $3 ; or
•ince 3 -f- 8 = f , we may say that the quotient is $1213f . Hence, 1
acre will cost $1213 J, when 8 acres cost $9707.
Note. — It is obvious that in getting l of the above number, we
have really divided it by 8.
(d.) By omitting the names of the denominations, we
ihould have —
I of 9 = 1, with 1 remaining.
J of 17 = 2, with 1 remaining.
i of 10 = 1, with 2 remaining.
i of 27 = 3, with 3 remaining.
This gives 1213, with 3 remaining, or 1213^, as before.
(c.) By only naming the quotient figures, we have —
1 thousand, 2 hundreds, 1 ten, 3 units, and 3 remainder; which
gives the same answer as before.
Proof. — If 1 acre costs $1213|, 8 acres will cost 8 times S1213| ; or,
which is the same thing, 8 times $1213, plus $3. This will be $9707,
if the work is correct.
J. 87. Decimal Fractions in the Quotient.
(a,) The answer to each of the above questions might
have been obtained in another form, by reducing the final
remainder to lower denominations.
(6.) In the first example, the 7 units remaining = 70 tenths, and 8 is
contained in 70 tenths 8 tenths times, with a remainder of 6 tenths.
But 6 tenths = 60 hundredths, and 8 is contained in 60 hundredths 7
hundredths times, with a remainder of 4 hundredths But 4 hundredth*
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DIVISION. 105
B= 40 thousandths, and 8 is contained in 40 thousandths 5 thousandthi
times. This V7ould give the following written work : —
Divisor S8 ) $58455.000 = Dividend.
7306.875 = Quotient = number of barrels.
{c.) In the second example, the 3 units remaining equal 30 tenths,
and |. of 30 tenths = 8 tenths, with a remainder of 6 tenths. But 6
tenths = 60 hundredths, and 1. of 60 hundredths = 7 hundredths, with
a remainder of 4 hundredths. But 4 hundredths = 40 thousandths, and
j of 40 thousandths = 5 thousandths. This would give the following V
BTitten work : —
Divisor = 8 ) $9707.000 = Dividend.
$1213.375 = Quotient = cost of 1 acre.
^d.) The answers in this form can be proved in precisely
tiie same way as were the former answers.
88. Compound Division.
(a.) All questions in compound division must of necessity
belong to the second class.
A benevolent society divided 10 cwt. 3 qr. 20 lb. of flour
equally among 6 poor persons. What was each person's
share ?
Reasoning Process. — If 6 persons had 10 cwt. 3qr. 201b. of flour, 1
person would have i of 10 cwt. 3 qr. 20 lb.
WRITTEN WORK,
cwt qr. lb.
6 ) 10 3 20 = share of 6 persons.
————— oz.
13 7 8 = share of 1 person.
Explanation of Process. 4 of 10 cwt. = 1 cwt., with a remainder of
I cwt. But 4 cwt. = 16 qr., and 3qr. added = 19qr. ^ of 19qr. =s
5 qr.. with a remainder of 1 qr. But I qr. = 25 lb., and 20 lb. added tm
i:» lb. ; ^ of 45 lb. = 7 lb., with a remainder of 3 lb. But 3 lb. = 48 oz.,
and X of 48 oz. = 8 oz.
llence, the quotient is 1 cwt. 3 qr. 7 lb. 8 oz.
Second Explanation. 6 is contained in 10 once, with 4 remainder
I cwt = 16 qr., and 3qr. added = 19qr. 6 is contained 3 times in 19,
with 1 remainder. 1 qr. = 25 lb., and 20 lb. added = 45 lb. 6 is con*
iained 7 times in 45, with 3 remainder. 3 lb. =» 48 oz., and is con*
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106 DIVISION.
taincd 8 times in 48 Hcocc, the quotient ifl 1 cwt 3qr. 7 lb. 80s., m
bflorc.
(b.) This may 1 3 proved as were tLe examples in simple
numbers.
Note. — Had the division been carried no farther than to pounds^
the answer would have been I cwt. 3 qr. 7 lb., with a remainder of 3 lb.,
or I cwt 3 qr. 7i lb.
S9* Problems for SoItttioUy including deduction Ascending^
"WTiat is the quotient —
1.
Of 87543 ~ 7 ?
6.
Of 348724 -T- 4?
2.
Of 59487 ~ 3 ?
7.
Of 37927 -T- 6?
8.
Of 38640 ~ 6 ?
8.
Of 29684 -1- 9 ?
4.
Of 827546 -T- 9?
9.
Of 20034 -j- 6?
5.
Of 217483 — 8?
10
. Of 7248396275436002(
)31
«
9?
11. How many yards of German broadcloth at $6 per
yard can be bought for $174564 ?
12. How many sheep at $7 apiece can be bought for
$7833 ?
13. If 8 melons cost $1, how many dollars will 3744
melons cost ?
14. When 6 books are bought for $1, how many dollars
must be paid for 1743 books ?
15. How many kegs, each holding 9 gallons, can be filled
fi-om 4752 gallons of molasses ?
16. How many hours will it take a vessel to sail 8937
miles, if she sails 8 miles per hour ?
17. How many clocks at $6 apiece can be bought for
$3528?
18. How many bags of coffee at $9 per bag can be bought
for $7487 ?
19. How many casks of raisins at $7 per cask can be
bought for $6594 ?
20. How many hours would it take a man who walks at
the rate of 4 miles per hour to walk 4739 miles ?
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DIVISION. 107
21. How many hats at $4: apiece can be bought for
$1372 ?
22. 31741 quarts =z how many bushtjls, pedLS, and quarts?
WRITTEN WORK.
8 ) 31741 — 5qt
4 ) 3967 — 3 pk.
991 bu.
Reasoning Process and partial Explanation, — Since 8qt ss Ipk^
31741 qt. must eqnal as many pecks as there are times 8 in 31741, which,
found by the usual method, gives 3967 pk. 5 qt. We now reduce the
pecks to bushels. Since 4 pk. = 1 bu., 3967 pk. must equal as many
bushels as there are times 4 in 3967, which, found by the usual method,
gives 991 bu. 3 pk. Hence, 31741 qt. = 991 bu. 3 pk. 5 qt
Note. — Problems like the above, in which it is required to find the
value of a number of units of one denomination in terms of some
higher, are ordinarily called Problems in Reduction Ascending; but they
do not differ in their nature from other problems in division.
23. 87633 days = how many weeks ?
24. 698472 inches = how many feet?
25. 98464 3 = how many ounces ?
26. 7987*^36 square feet = how many square yards ?
27. 97875 quarters = how many yards and quarters r
28. 87637 quarts = how many bushels, pecks, and quarts?
29. 798953 gills = how many gal., qt., pt, and gL ?
30. 793527 9 = how many lb, S , 3, and 9 ?
31. 587537 qr. = how many £, s., d., and qr. ?
32. 23975 fur. == how many le., m., and fur. ?
33. 15951 na. = how many yd., qr., and na. ?
34. 63359 pt = how many bu., pk., qt, and pt. ?
S5. 57984 gi. = how many gal., qt., pt, and gi. ?
36. 7951 qr. = how many £., s., d., and qr. ? *
37. 4375 3 = how many lb, S , 3, and 3 ?
38. A farmer put 75 gal. 3 qt 1 pt 2 gi. of cider into
bottles, each contwning 1 pt. 2 gi. How many bottles did it
take to contain it ?
Reasoning Process. — Since it takes 1 bottle to hold 1 pt. 2 gi., it must
take as many bottles to hold 75 gal. 3 qt 1 pt 2 gi. as there are timet
1 pt 2 gi. in 75 gal. 3 qt 1 pt 2 gi.
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108 DIVISION.
Note. — Before performing the division, each quantity most Dm
reduced to gills.
39. A grain dealer put 498 bu. 3 pk. of grain into bags,
each holding 1 bu. 3 pk. How many bags did he fill ?
40. I bought 536 yd. 2 qr. 2 na. of tape, which I cut into
pieces 2 qr. 1 na. long. How many pieces did it make ?
41. If 9 men earn $4375.26 in a year, how much will 1
man earn in the same time ?
Bmsoning Process. — If 9 men cam $4375.26 in a year, 1 man will earn
1 of $4375.26 in the same time, which may he foand by dividing by 9.
42. If 7 cases of cloth cost $6545, how many dollars will
1 case cost ?
43. If 18579 lb. of hay are obtained from 9 acres, how
many pounds are obtained from 1 acre ?
44. If 7 horses can draw 16786 lb., how many pounds can
1 horse draw ?
45. If it cost $173549 to build 9 miles of railroad, how
much will it cost to build 1 mile ?
46. A father divided his estate, valued at $879643, equally
among his 5 children. What was the share of each ?
47. What is f of 975646 ?
48. What is \ of 39547 bushels ?
49. What is \ of 7354.278 tons ?
50. What is \ of 12456.78 miles ?
51. If 6 piano-fortes cost $1837.44, how much will 7 cost ?
Reasoning Process. — If 6 piano-fortes cost $1837.44, 1 will cost ^
of $1837.44, which (found by dividing by 6) is $306.24. If 1 piano-forti
costs $306.24, 7 will cost 7 times $306.24, which (found by multiplying
by 7) is $2143.68.
52. If 9 wagons cost $1378.71, what will 4 cost?
53. If 8 gold watches cost $775.36, what will 9 cost?
54. K 1 acre of land costs $327.52, what will 3 roodd
. cost ?
bb. What is the cost of 5 cord feet of wood at $7,376 per
cord ?
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DIVISION. l09
56. How far will a man travel in 6 days, if he travels at
die rate of 174 79 miles per week?
57. If 1 bushel contains 2150.4 cubic inches, how many
cubic inches will 3 pecks contain ?
58. If 1 peck contains 537.6 cubic inches, how msiny cubic
inches will a gix-quart basket contain ?
59. If in 1 quart, liquid measure, there are 57.75 cubic
inches, how many cubic inches are there in 1 qt 1 gi., or 9
gills?
60. If 1 man can perform a piece of work in 2944 hours,
in how many hours can 8 men perform the same work ?
Note. — It is obvious that 8 men will perform it in i. of the number
of hours required by 1 man.
61. If it takes 1 laborer 3474 days to earn $5211, how
many days will it take 6 laborers to earn the same sum ?
62. In a certain fort there are provisions enough to keep
1 company of 76 soldiers 7256 days. How long can 8 com-
panies of the same size be kept on them ?
63. 10 men bought a stack of hay weighing 4 T. 16 cwfc.
qr. 1 lb. 14 oz. 6 dr. What ought each man's share *o
weigh ?
64. A farmer had 9 equal bins filled with grain. They all
contained 2047 cu. fl. 1436 cu. in. How many feet and inches
did each bin contain ?
65. The brig Maria sailed from Philadelphia to Boston,
with 207 T. 13 cwt. 2 qr. 19 lb. of coal, | of which was land-
ed at one wharf, and the remainder at another. What was
the weight of that landed at the first wharf? at the second?
66. K John walks ^ as fast as George, how far will John
walk while George is walking 97 m. 6 fur. 38 rd. ?
67. A company of 9 California gold diggers found in 1
month 371b. 4oz. 16dwt. 17 gr. of gold, which they divided
equally. What was the share of each ?
. 68. 8 men bought 17 T. 18 cwt. 3 qr. 161b. of sugar,
which they divided equally. What was the share of each
manT •
10
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110 DIVISION.
69. If 7 cowdi eat ICT. 5 cwt. 1 qr. 181b. of hay in 1
year, Low much will D oowr? eat in the same time?
70. If 6 equal pieces oi" cloth contain 185 yd. 2 qr. 2 Da«
!iow much will 5 pieces of ihe same size contain ?
71. it 5 pieces of broadcloth cost £87 13 8, 9 d., bow
much will 7 pieces cost ?
72. What is ^ of 374 lb. 11 oz. 10 dwt- 20 gr. ?
73. What is i of 857 m. 5 fur. 86 rd. 4 yd. 2 ft. 6 in. ?
74. What is ^ of 17 yr. 135 da. 14 h. 37 m. 56 sec. ?
75. What is i of 15*' 35' 54''?
76. What is i of 191b 115 63 29 15 gr.?
90. Division hy Factors,
(a.) When the divisor is the product of factors, it will
sometimes be convenient to divide by its factors, instead of
dividing by the whole number at once.
(6.) To understand the process of division, and method of getting
the true remainder, reduce 15262 quiirts to busliels and quarts, by first
reducing it to bushels, pecks, and quarts, as before explained, and com-
pare the explanation and work with the explanation and work of the
following example.
(c.) What is the quotient of 15262 divided by 32 ?
WRITTEN WORK.
8 ) 15262 — 6, the first remainder.
4 ) 1907 times 8,-3 times 8, or 24, the 2d remainder.
476 times 32 = quotient sought.
24 -|- 6 = 30 = true remainder.
Explanation, — Since 32 = 4 times 8, we may divide by 8, and then
by 4. Dividing by 8 gives 1907 for a quotient and 6 for a remainder,
i. e., 1907 times 8, or parts of 8 each, with 6 units remaining undivided.
Dividing 1907 by 4 gives 476 for a quotient and 3 for a remainder,
I. e., 476 parts each equal to 4 of the former ones, or to 4 times 8, of
32, with 3 times 8, or 24, undivided. Adding 24 to the former remain-
der gives 30 for the true remainder. Hence, 15262 = 476 times 32,
with a remainder of 30, or it equals 4763.3. times 32.
{d.) Had the problem been to find ^L. of 15262, the first' quotient
would have represented the number of units in each part obtained )t •
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DIVISION. in
dividing 15262 into 8 equal pnrts; and the second the nnmber of anits
in each part obtained by dividing each of these 8 parts into 4 others
or, which is the same thing, the numl»cr of units there would be in ca^
part obtained by dividing 15262 into 32 equal parts.
To get the true remainder, observe that as uacli of the 8 parts gi^ei
a remainder of 3 units, all of them must give a i*eraainder o^ 8 times 3,
or 24, which, added to the 6 left by first division, gives 30 for tho true
remainder.
The work would be written tbus : —
8 ) 15262 — 6 first remainder.
4 ) 1907 units in each of 8 parts — 3 remainder on
[each part, or 8 times 3 in all.
476 units in each of 32 parts.
24 -|- 6 = true remainder.
(c.) From the above, it is evident that the true quotient
will be obtained by dividing the dividend by one factor, the
quotient of this division by another, the quotient of the last
division by another, and so on, till all the factors of the set
considered are used.
{/.) The true remainder may be obtained by multiplying
the remainder of each division by the divisors of all the pre-
ceding divisions, and adding the products to the remainder of
the first division.
{g.) What is the value —
1.
2.
3.
4.
5.
(Ji.) The most important application of the division by
factors is made when the divisor is a multiple of 10, or of
eome power of 10. In such cases we first divide by the
power of 10, and then that quotient by the other factor of the
divisor, getting the true remainder as before explained.
11. What is the quotient of 578635 -7- 800 ?
Explanation. — Dividing by 100, by removing the point two places to
tho left, gives 5786 for a quotient and 35 for a remainder, and dividing
Of 74385 ~ 49 ?
6.
Of 57482 -^. 15 ?
Of 67849 -^ 35 ?
7.
Of 28654 -^ 21 ?
Of 4535 ~ 24 ?
8.
Of 5477 -^ 18?
Of ^V 0^ ^^358 ?
9.
Of ^V of 20249 ?
Of 2V 0^ 3^4'^ ?
10.
Of Jff of 172783
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112 '♦IV1810N.
this quotient by 8. gives a final quotient of 723 and a rcmaiBder of t
times 100, or 200, which, added to 35, the first remainder, gires 235 at
the true remainder.
The work may be written thus : —
100 ) 578635 — 35
8 ) 5786 — 2
'723
[=2X
100 + 35 = 235.
(t .) What is the quotient —
12. Of 6379 — 40?
18. Of 27476 -J- 300 ?
14. Of 427875 ~ 9000 ?
15. Of 258647 -7- 80000 ?
16. Of 174326-7-50?
17. Of 927673 -T- 7000 ?
18. Of 42700-1-9000?
19. Of 23750000-7-30000?
91 • Divisor a large Number. — Trial Divisor,
(a.) When the divisor is a large number, it is not alwajs
easy to tell at once what is the true quotient figure. In such
cases, the number expressed by one or two of the left; hand
figures of the divisor may be selected as a sort of trial divi^
sor ; but to determine whether the quotient figure thus ob-
tained is the true one, it will be necessary to find the product
of the divisor by the quotient, and the remainder after sub-
tracting this product from the dividend.
(5.) The product of the divisor by the quotient should
either be equal to, or less than, the dividend. When it is
equal to the dividend, the division can be exactly performed :
but when it is less, there is a remainder, which may be found
by subtracting it from the dividend.
{c) If, in any case, the product of the diviso:* by the sup-
posed quotient is greater than the dividend, it shows that the
divisor is not contained as many times in the dividend as the
supposed quotient indicates.
{d,) We will now apply these principles to a few examples.
1. What is the quotient of 387 -i- 43 ?
Solution. — Since 43 differs but little from 4 tens, we may infer that'
the entire part of tJie quotient of 387 -s- 43 will be nearly or precisclj
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DIVISION. lis
tne same as that of 38 -I- 4, which is 9. To ascertain whethc? rfiis be
correct, we nuiltipljr 43 by it, which gives exactly 387, and proTes our
work.
The figures may be written thus : —
Divid.
Divisor = 43 ) 387 ( 9 = Quotient
387 = 9 times 43.
2. What is the quotient of 6169 -^ 825 ?
Solution. — Since 825 differs but little from 8 hundi*eds, we may infer
that the entire part of the quotient of 6169 -5- 825 will be nearly or pre-
cisely the same as that of 61 -*- 8, which is 7. To ascertain if this is
correct, we multiply 825 by 7, which gives 5775 for a product This be-
ing less then the dividend shows that the quotient figure is not too
large* Subtracting this product from the dividend gives 394 for a
remainder, which, being less than the divisor, shows that the quotient
figure is correctt
Hence, the quotient is 7, and the remainder is 394, or the complete
quotient is 7^^ 4.
WRITTEN WORK.
Divid.
Divisor = 825 ) 6169 ( 7 = Quotient
5775 = 7 times 825.
394 = Remainder.
3, What is the quotient of 55673 -!- 6349 ?
Solution. — Making 6 the trial divisor, we have 9 for a tnal quotient.
Multiplying 6349 by 9 gives 57141 for a product, which, being greater
than the dividend, shows that the quotient figure is too large. Select-
ing 8 as the quotient figure, and multiplying as before, gives 50792 for
a product, which, being less than the dividend, shows that 8 is not too
large. Subtracting 50792 from 55673 gives a remainder of 4881.
Hence the quotient is 8, and the remainder is 4881, or the complete
quotient is 8J§|J.
WRITTEN WORK.
Divid.
Divisor = 6349 ) 55673 ( 8 = Quotient
50792 = 8 times 6349
4881 = Kemainder.
• For it shows that the divisor taken 7 times is less than the dividend.
t For it shows that if the divisor had been taken once more, the result
W*>uld have been more than 6169.
10 ♦
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}14 DIVISION.
4. "WTiat is the quotient of 3913 -^- 482 ?
Solution. — Since the divisor U nearer to 5 hundreds than 4 hnudrcdd
we select 5 as the trial divisor. This gives 7 for a trial qaoticnt, and
multiplying giws 33*4 for a product, which, being less than the divi
dend, shows that 7 is not too large. Subtracting 3374 from 3913 gives
a remainder of 539, which, being greater than the divisor, shows that 7
is too small. Substituting 8 in its place, and multiplying and subii act-
ing as usual, gives the following written work —
482 ) 3913 (3
3856 = 8 X 482.
57 = Remainder.
Hence, the quotient is 8, and the remainder is 57, or the complete
quotient is 8^5^^
What is the quotient —
5. Of 5279 ~ 847 ?
6. Of 19476-4-2784?
7. Of 35947 4-8912?
11. What is the quotient of 15078 -r 276 ?
WRITTEN WORK.
8. Of 394087 — 182578?
9. Of 807436 -r 294367 ?
10. Of 42974 -^ 8523 ?
276 ) 15078 ( 54
1380 = 50 X 276
1278
1104 = 4 X 276
174 = Remainder.
Solution. — Making 3 the trial divisor, we may infer that as 3 is con-
tained 5 times in 15, 276 must be contained 5 tens times in 1507 tens.
Writing 5 as the tens' figure of the quotient, we multiply the divisor by
it, and place the product, which is 1380, under the tens of the dividend,
to show that its denomination is tens. Subtracting this from 1507 tens
leaves a remainder of 127 tens, to which adding the 8 units gives 1278
units to be divided.
Since 3 is contained 4 times in 12, we may infer that 276 is contained
t times iij 1278. Writing 4 as the units' figure of the quotient, we mul-
tiply the divisor by it, and write the product, which is 1104, under the
dividend. Subtracting leaves a remainder of 174, which is less than
the divisor.
As we have now considered all the denominations of the dividend,
•V6 may consider the division complete, (unless wc wish to have th«
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Divisioy 115
quotient appear in the form of a decimal tractKn.) Hence, the qaot«ji|
is 54, and the remainder is 174, or the «nplcte quotient is 541X4.
Note. — It is obvious that in pcr».-^:»'(; the above work, we suV
^ tracted 50 times the divisor, and t(^^ 4 tiroes the divisor, which is
equivalent to 54 times the divisor, from the dividend, and that this left a
remainder of 174.
(c.) Had it been desirable to have the answer contain a
d<)cimal fraction, we should have continued the division, thus:—
174 units = 1740 tenths, and as 3 is contained 5 times in 17, we ma>
infer that 276 is contained 5 tenths times in 1740 tenths. Multiplying
the divisor by 5, and subtracting the product as before, gives a remain-
der of 360, which, being greater than the divisor, shows that the divis'-»r
would have been contained more than 5 tenths times in the dividen'i.
We therefore substitute 6 for 5 in the tenths place of the quotient, a^id
having erased the last product and remainder, multiply the divisor by 6,
subtracting the product as before. This gives a remainder of 84 temhs,
which may be reduced to hundredths, and divided as already explained*
In the following form, the division Is carried out to mUlionllitt.
276 ) 15078. ( 54.630434
1380 = 5 tens X 276
1278.
1104. = 4 times 276
174.0
165.6 = 6 tenths X 276
8.40
8.28 = 3 hundredths X 276
.1200
.1104 = 4 ten-thousandths X 276
960 = hundred-thousandths
828 = 3 hundred thousandths X 276.
1320 = millionths
1104 = 1 millionth X 276
216 = millionths = Hcmainder.
NoTK. — It is obviDus that we have subtracted from the dividend the
Products of the divisor multiplied by 5 tens, by 4 units, by 6 tenths, by
$ hundredths, by 4 ten-thousandths, by 3 hun Ircd-thousan.dths, and by I
millionth ; and that this leaves a ren.ainde^ of 44 millionths. Henco^
if the work be correct the sum of these se eral products added to the
final remainder will equal the dividend. The methods of proof befor*
explained will also apply.
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ill DIVISION.
9!2« Long Division,
(a.) When, as in 8«5 to 91, we write only the dins&r,
dividend, quotient, and final remainder, the prcKjess is called
Short Division; but when, as in 91, we write the divisor,
dividend, and quotient, and also the products of the divisor by
the successive figures of the quotient, together with the re-
mainders obtained by subtracting these several products from
the corresponding denominations of the dividend, the process
is called Long Division.
(J>.) Long Division differs from Short Division in these
respects, viz.: First That having obtained a quotient fig-
ure, we multiply the divisor by it, writing the product under
the part of the dividend considered.
Second. That we subtract this product from the paii of the
dividend considered, writing the remainder.
Third. That we write the next figure of the dividend after
this remainder, to form the next partial dividend.
(c.) The real difference between them may be stated thus :
Li long division more of the work is written than in short
division.
{d.) Long division is generally employed when the divisor
is a large number.
(c.) If at any time the product of the divisor by a quotient
figure should be less than the part of the dividend considered,
it would show that the quotient figure was too large. (See
Sd example, 91.)
(/.) If the remainder obtained by subtracting from the
dividend the product of the divisor by any quotient figure
should be greater than the divisor, it would show that the
quotient figure was too small. (See 4th example, 91 •)
ig.) Perform the following examples by long division.
See solution to 11th example in 91.)
What is the quotient —
1. Of 3784 -T- 21 ?
2. Of 59378 -T- 82 ?
8. Of 386495 -^ 289 ?
'4. Of 43625 -^714?
5.
Of 386427 -^ 5287 ?
0.
Of 2700684 -T- 19743 ?
7.
Of 438277 4-34?
8.
Of 6293876 ~ 2874 ?
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DIVlSirN. 11/
9. How many square feet are eqaal to 42912 sqoara
inches ?
10. How many boxes, each containing 73 quarts, can be
filled from 658764 quarts of meal?
11. How many watches at $48 apiece can be bought for
$3456 ?
12. If a boy can perform 37 examples in one day, how
many days will it take him to perfonn 8764 examples ?
13. If a vessel sails 147 miles per day, how many days
will it take her to sail 3579 miles ?
14. If 252 pints of cider will fill a barrel, how many baiv
rels can be filled from 876438 pints ?
15. How many weights, each weighing 56 lb., will be re-
quired to balance 7 loads of iron, each weighing 9472 lb. ?
16. James can pump 37 quarts of water per minute, and
William can pump 43 quarts of water per minute. How
many minutes will it take William to pump as much as James
can pump in 473 minutes ?
17. What will 1 acre of land cost if 237 acres cost
$236526?
18. If 687 bushels of potatoes cost $515.25, what will 236
bushels cost ?
19. If 67 yd. of silk cost $75,375, what will 48 yd. cost?
20. If 89 barrels of flour contain 17444 pounds, how
many pounds wiU 258 barrels contain ?
93. Abbreviated Process.
(a,) By performing the subtraction at the same time that
we do the multiplication, much of the written work can be
avoided.
(5.) The subtraction may be performed in accordance with
the principles explained in 73. The following example
will illustrate it. We have, however, given only the mechan-
ical process, leaving it with the pupil to explain the various
steps.
1. What is the quotient of 1947.63 -7- 396 ?
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.^8 DI/ISION\
Procesi, 396 is contained 4 lirncs in 1947. 4 times 6 = 24, which
sahtracteJ from 27 leaves 3 ; this wc write in tlie units* place. 4 times
9 = 36, and 2 added make 38, which subtracted from 44 leaves 6. 4
times 3 = 12, and 4 added make 16, which subtracted from 19 leaves 3.
The first remainder is, therefore, 363. Reducing this to tenths, and
adding the 6 tenths, wc have 3636 tenths, which divided by 396 gives 9
tenths for a quotient. We multiply and subtract as before, and thus
proceed till the division is completed, or till we have as many decimal
J laces in the qu'^ticnt as wc wish.
WKITTEN WORK.
896 ) 1947.63 (4.9182 = Quotient
363.6
7.23
3.270
.1020
.0228 =: Remainder.
(c.) Perform the division in the following examples in the
same way : —
2. What is the quotient of 78643 -^ 47 ?
3. What is the quotient of 137648 -r 326 ?
4. What is ^V ^^ ^5427 ?
5. What is ^7 of 3865726 ?
6. What is the value of 764379 cubic inches, expressed in
cubic feet and inches ?
7. How many square feet and inches are equal to 954376
square inches ^
8. If the wages of 96 laborers for 1 year are S26470.08,
what will be the wages of 1 laborer for the same time ?
9. If 875 pairs of boots cost $2737.16, how much will 1
pair cost ?
10. How many pairs of shoes at $2.37 per pair ciin be
bought for $184.86 ?
Note. — In solving such examples as the above, both divi?or and
dividend should be reduced to cents, thus : if for 237 cents one pair of
shoes can bo bought, for 1S486 cents as many pairs can be bought as
there are times 237 in 18486. which may be found by methods already
explained.
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CONTRACTIONS AND 3IISC1:LLANE0C8 PROBLEMS. 119
11. How many pounds of tea at $.47 per pound can bo
bought for $464.23 ?
12. How many busliels of wheat at $1.85 per bushel can
be bought for $287.35 ?
13. How many pounds of raisins at $.125 per pound can
be bought for $39,875 ?
14. How many days must a man labor at $1 75 per day
to earn $596.75 ?
15. I gave 17 acres of land worth $46.98 per acre in ex-
change for wild land at $4.59 per acre. How many acres of
wild land did I receive ?
16. 1475869 cu. in. = how many C, Cd. ft., cu. ft,, and
cu. in. ?
17. 8383794 oz. = how many T., cwt, qr., lb., oz., and dr. ?
18. 67458 sq. rd. = how many A., R., and sq. rd. ?
19. 57864 gr. = how many lb., oz., dwt., and gr. ?
20. 457986 cu. in. = how many C, Cd. ft^, cu. ft., and
cu. in. ?
SECTION VIII.
CONTRACTIONS AND MISCELLANEOUS PROB-
LEMS.
Introductory Note. — It is frequently the case that by carefully exam-
icing the numbers we are to operate tfpon, we can discover some abbre-
Tiated method of performing the work. We have already suggested
several such methods, and it is the design of this section to suggest
othors.
•4. Multiplier a convenient fractional Part of 10, 100
1000, Sfc.
When the multiplier is a convenient fractional part of 10,
too, 1000, or a unit of any higher denomination, the follow-
mg principles may be advantageously applied.
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120 OONTBACTIOXS AND MISCEI.LAXEOUS PBOBLILMS.
1. How many are 25 times 679 ?
Solution. — Since 25 equals one fourth of 100, 25 times 679 muit
Aqual one fourth of 100 times 679, or ± of 67900, which equals 1697.^
2. How many are 25 times 657 ?
8. How many are 50 times 657 ?
4. How many are 12^ times 834 ? (12^ = i of 100.)
5. How many are 16J times 957 ? (16$ = ^ o/ 100.)
6. How many are 33^ times 871 ? (33^ = i of 100.)
7. How many are 14f times 249 ? (14f = + o/ 100.)
8. How many are 3^ times 249 ? (8J = ^ of 10.)
9. How many are 2^ times 822 ? (2^ = ^ of 10.)
10. How many are 6f times 944? (6^ = ^ of 100.)
11. How many are 333^ times 9478? (333^ = i of
1000.)
12. How many are 125 times 8767 ? (125 = i of 1000.)
18. How many are 250 times 6894 ? (250 = i of 1000.)
9Sm One Part of Multiplier a Factor of another Part.
When one part of the multiplier is a factor of another part^
much labor can often be saved by applying the principle*
illustrated in the following examples : —
1. What is the product of 7678 multiplied by 427 ?
Solution, — By examining the multiplier, we perceive that it equa>M
420 + 7 = 42 tens + 7 units, and that 42 tens = 6 tens or 60 timsK
7. Hence, we have the following written work.
a= 7678
b = 427
7 times a = c = 63746
6 tens, or 60 times c s= 420 times a == d == 322476
c + d =5 b tunes a = e = 3278506 = Ais.
2. What is the product of 72144 times 874369 ?
Explanation, — By examining the multiplier, we perceive that it
equals 72 thousands + 144, and that as 144 = twice 72, it must equal
S times 001 of 72 thousands. Hence.
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OONTKACT.ONS AND MISCELLANEOUS PROBLEMS. 121
a r= 874369
b = 72144
72000 X a = c « 62954568000
2 Umes .001 of c = 144 times a = d = 125909136
c + d = b times a =» e = 63080477136
8. How many are 1875625125 times 97643721785 ?
a = 97643721785
b » 1875625125
125 times a = c = 12205465223125
3000 times c = 625000 times a =» d = 61027326115625
SOOO times d = 1875000000 times a =» e =» 183081978346875
c + d + e = b times a = 188143017878455848125
How many are —
4. 14874 times 876437?
5. 19899 times 879438 ?
6. 17525 times 43678 ?
7. 83415 times 476437 ?
8. 43821973 times 976327864?
9. 36324 times 5428^732 ?
10. 1998999 times 463829748 ?
11. 1752512^ times 83748954?
96. To divide by 99, 999, ^c.
(a.) Since 10 = 9 + 1, and 100 = 99 + 1, and 1000
= 999 + 1, &c, it follows that 40 = 4 times 9 + 4; that
2800 = 28 X 99 + 28 ; that 37900 = 379 X 99 + 379 ;
that 786000 =r 786 times 999 + 786; &c On this princi-
ple the following processes are based.
1. What is the quotient of 437 -7- 99 ?
Solution. 437 = 400 + 37; but 400 = 4 times 99 + 4. Hence,
137 = 4 times 99 + 4 + 37 = 4 times 99 + 41 = 4|J times 99 =»
Anstver,
2. What is the quotient of 15378 -r 99 ?
S(^ut{on. 15378 = 15300 + 78 = 153 times 99 + 153 + 78 =3
153 times 99 + 231. But 231 = 200 + 31 == 2 times 99 + 2 + 31
11
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122 OONTBAOTIONS AND MISCELLANEOUS PROBLBMS. >
■B 2 times 99 + 83. Now, by adding 153 times 99 to 2 times 99, w»
hare 155 times 99. Hence, 15378 = 155 times 99 + 33 = 15l||
times 99.
The following exhibits a convenient form for writing the work : —
153
2
78
31
155 33=155J| .
The fall explanation is the same as that already given The neces*
gary numerical operations are as follows: Separating the hundreds
from the tens and units by a vertical linei '^o ^^d ^^^ number at the
right of the line to that at the left This gives 153 + 78 = 231 =£r8t
remainder. Writing this beneath the dividend, and adding as before,,
we have 2 + 31 = 33, which, being less than 99, is the final remainder.
The sum of the uumbers at the left of the line, or 153 + 2 = 155, the
quotient.
(k) In a similar manner we can divide by 999, 9999, &c.
f#) Let the pupil find, if he can, the application of a
similar principle to the division by 98, 97,, 96, ^c. ; also to
998, 997, &C., and afterwards perform the following exam-
pies : —
What is the quotient —
8. Of 186738 -f- 99 ? i
4. Of 49763842 -1- 999 ? J
5. Of 763852748 -r- 9999 ?
6. Of 9842987483 -7- 99999?
7. Of 54783 4- §8?
8. Of 2987637 — 96? -
9. Of 248763 -i- 997?
10. Of 69874325 4- 9998 ? . .
97, To divide hy any convenient fractional FaH of 10, ,
100, or 1000.
(a.) Since 100 -r 25 = 4, 900 ^ 25 must equal 9 times
4; 1700 -^ 25 must equal 17 times 4; 49600 -f- 25 must
equal 496 times 4.
(h.) Since 100 — 12J = 8, 3800 -J- 12^^ must equal 38
limes 8 ; 49700 -^ 12^ must equal 497 times 8, &c
(c.) Since 1000 -M25 = 8, 479000 -f. 125 must equal
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CONTRACTIONS AND MISCELLANEOUS PH0BLEH8. 123
479 times 8 ; 3785000 -— 125 must equal 3785 times 8,
(rf.) These principles are applied in the following pro-
cesses.
1. What is the quotient of 9738 -i- 25 ?
Soiutiotu 9738, 9700 -|- 38. fiat 9700 = 97 times 100 => 97 timet
• 4 times 25 = 388 times 25 ; and 38 =: 1 J.^ times 25. HencCi 9738 -••
25 = 383 + 1 Jl == 389^|.
What is the quotient —
2. Of 86794 -i- 25 ?
8. Of 3475 -M2^?
4. Of 6950 -r- 16^ ?
5. Of 42725 -r- 6^ ?
6. Of 54737 -J. 125 ?
7. Of 98734 -7- 33^?
8. Of 4765465 -i. 333^ ?
9. Of 657847^250?
98* Miscellaneous Problems.
1. A man bought a house lot and garden for $1378.24;
he paid $4796.87 for building a house, $1274.38 for building
a stable and carriage house, $438.47 for fencing the lot,
$578.37 for laying out the garden and grounds, $1287.63
for a span of horses and a carriage, $1328.56 for furnishing
his house, and then had $47289.43 left. How much monej
did he have at first ?
2. Mr. French bought a house for $6742.38, and after pay-
ing $138.47 for having it painted, and $527.94 for repairs,
he sold it for $8472. Did he gain or lose, and how much ?
3. Mr. Hall bought 437 cords of wood at $3 per cord, and
sold it at $4.75 per cord. What did he gain by the specula-
tion ?
4. A man started on a journey of 1 1 64 miles, and trav-
elled 37 miles per day for the first 21 days. How many days
would it take him to finish the journey if he should travel at
the rate of 43 miles per day ?
5. Bought 43 acres of land at $28.73 per acre, and sold it
at $37.73 per acre. How much did I gain ?
6. I bought 487 yards of cloth at $2 per yard, and 627
yards at $4 per yard ; I sold the whole of it at $4 per yard
How much did I gain ?
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124 CONTBACTIONS AND MISCELLANEOUS PROBLEMS.
7. A trader bought 528 barrels of flour at $7 per barrel,
875 barrels at $8 per barrel, and 497 barrels at $9 per bar
rel. He sold the whole of it at $8 per barrel. Did he gain
or lose, and how much ?
8. How much will a house lot, 137 feet long and 89 feot
wide, cost, at 2 cents per square foot ?
9. Which is the larger 13 X 17 X 28 X 43, or 28 X 13
X 43 X 17, and how much ?
10. A man who had $4376 invested it in flour at $8 per
barreL He sold 238 barrels of the flour at $9 per barrel,
and the rest for enough to make up $5232. For how much
did he sell the last lot per barrel ?
11. A man bought a horse for $237 ; he kept him 19
weeks at an expense of $2.50 per week, and then exchanged
him for another horse, receiving $48 for their difference in
Talue. After keeping the second horse 4 weeks, at an ex-
pense of $2.25 per week, he sold him for $225. Now, allow
ing that the use of each horse was worth $1.75 per week, did
he gain or lose by the transaction, and how much ?
12. To 12 times 487 add 8 times 683, multiply the sum by
7, and subtract 4279 from the product
13. James Smith bought of John Brown 7 hogsheads of
molasses, each containing 147 gallons, at $.27 per gallon, and
gave in payment $100 in money, and the rest in nails at 5
cents per pound. How many pounds of nails did it take ?
14. How many cords of wood at $4 per cord can be bought
for 82 barrels of apples at $2 per barrel ?
15. John earns $4 per week, and William earns $7. How
many weeks will it take John to earn as much as William can
^arn in 48 weeks ?
16. Gave 3 hogsheads of oil, each containing 167 gallons,
worth $1.68 per gallon, and $200 in money, for 4 acres of
land. How much ought the land to be worth per acre, that
I may neither gain nor lose ?
17. George had 378 oranges, which sold at 2 cents apiece.
With the money thus received he bought a box of oranges,
whicli he found contained 427. He ate 9 of these, gave away
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CONTRACTIONS AND atlSdELLANEOUS PROBLEMS. 125
7, had 4 stolen from him, sold 294 at 3 cents apiece, and the
rest at 2 cents apiece. Did he gain or lose on the box, and
• how much?
18. How many times will a carriage wheel. 12 feet 6 inches
,in circumference, revolve in going 5 miles, 4 fur., 18 rd., 4 yd.,
1ft., 1 in.?
19. A boy, riding with" his father, ascertained that the
hinder wheel of the carriage, which was 10 feet 4 inches iit
circumference, tevblved 1297 times in passing from one vil-
lage to another. How far apart were the villages, reckoning
;the distance in miles, furlongs, rods, &C.?
20. Mr Clarke's house lot is 83 feet wide, and contains
8051 square feet. How long is it ?
21. Mr. Angell says that his house lot is 97 feet long, but
that if it were 100 feet long, it would contain 168 square feet
more than it now does. How many feet does it now con-
tain?
' 22. Multiply 837 by 5, add 247 to the result, divide this
by 4, and calling the result dollars, find how many yards of
cloth at $2 per yard you could buy with it.
23. By buying a cargo of coal at $6 per ton, and selling it
at $8 per ton, I gained $198. How much did I pay for it?
24. A silversmith bought 13 lb., 4 oz., 2 dwt., 5 gr. of sil-
ver,. and after mixing with it 1 lb., 5 bz., 15 dwt., 19 gr. of
alloy, made it into spoons, each wdghing 1 oz., 4 dwt, 17 gr.
How many spoons did he make ?
25. I bought the wood standing on 9 acres of land, paying
for it at the rate of $67.25 per acre. I paid $.625 per cord
for having it cot, and $.75 per cord to have it carted to a rail-
road depoty where I sold it for $4.50 per cord. If there was
igi ^^yerage of 30 cords to the acre, how much did I gain by
the transaction ?
26. A trader .mixed 536 lb. of sugar, worth 9 cents per
pound, with 52 lb., worth 6 cents per pound, and 432 lb.,
worth 7 cents per pound. For how much per pound ought
lie to sell the mixture so s^s n^it))er to gain nor lose ?
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126 BILLS OF GOODS.
27. A trader bought 597 gallons of vinegar at 14 cents per
gallon, and after mixing with it 24 gallons of water, sold it
for 15 cents per gallon. How much did he gain by the trans«
action ?
28. I bought a pile of wood, 144 feet long, 4 feet wide,
and 6 feet high, at $3.92 per cord. What did it cost me ?
29. How many square feet in the walls of a room 18 feet
long, 1 6 feet wide, and 12 feet high ?
80. If William walks at the rate of 16 rods per minute,
and Joseph at the rate of 19 rods per minute, how long will
it take Joseph to overtake William, when William has 12
minutes the start
99. Bills of Goods.
(a.) When a man sells goods, he usually gives the pur-
chaser a written statement of the articles bought, and the
prices he is to pay for them. Such a statement is called a
« Bill of the Goods," or simply a « Bill."
(h,) A bill, like every other business paper, should contain
the date, i. e., the time and place of the transaction, and also
the names of the parties, and an account of the transaction.
If the goods are paid for, the bill should be receipted ; i. e.,
the words deceived Payment being written at the bottom, the
seller should affix his name.
(c.) The following example will illustrate this : —
Mr. George W. Dodge is a trader, residing in Lancaster.
On the 1st of January, 1855, he sold to Mr. Humphrey Bar-
rett, for cash, 3 yards of broadcloth at $3.75 per yard, 6 yards
of doeskin at $1.87 per yard, 32 yards of sheeting at 11 cents
per yard, 9 yards of black silk at 97 cents per yard, and 6
Hnen handkerchiefs at 34 cents apiece.
Mr. Dodge made out the bill as follows : —
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BILLS OF GOODS. 127
^Lanca^^let, Ian. / , /o55
3 yd,. 3^toaMd ae ^3-75 . 0//.S5
6 " Q^ocadin " /.§7 . //.SS
^ " e^Ac4 6f^J4" .^7 . B.73
6 " ^nenJ^Jly" .34 . 2.04
^36.76
(d.) If Mr. F. W. Spofford, Mr. Dodge's clerk, had re-
ceived the money due on the bill, he would have receipted it
by some form similar to the following : —
Or,
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128 BILLS OF GOODS;
(e,) If the goods had been bought on credit, the words
** Received Payment " might have been written, but no name
would have been affixed. It is the signing of the receipt by
the seller, or his authorized agent, which gives validity to it.
The receipted bill should be kept by the person who pays ity
as evidence of the payment,
(/.) The following form of heading a bill is often used
instead of the above : —
«^j^^vw^«^ ^^^» /, /q55.
Note. — A person is myMtor when he owes ihe money, and my
eredttor when I owe him.
(g,) When articles are bought or services rendered at dif-
ferent times, the bill may be headed and receipted as before,
and the dates of the various transactions written at the left,
opposite the entries.
(h,) Theodore Gay is a trader living in Deidham. He
sold to Samuel French the following article?, viz. : Jan. 3,
1855, 5 bags of meal at $1.58 per bag, and 2 bags of corn at
$1.54 per bag; Jan. 27, 8 lbs. of coffee at 16 cents per
pound; Feb. 7, 3 lbs. tea at 59 cents per pound; Feb. 21,
1 bbl. of flour for $10.37, and 14 lbs. of brown sugar at
9 cents per pound. On the 1st of March, Mr. Gay being in
want of money, made out a biU, and sent it to Mr. French^
who paid it on the 3d of March.
(u) The bill was marie out in the following form : —
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BILLS OP GOODS. 129
Jiin. 3, ^oz 5 ^a^^ ^ea/,at^^.6B, ^7-90
^J. 7, " 3 ^4. 5^ea, " .59, /.77
"S/, " ^m.M>m, " /0.37
" '' '' /^^ S^yat, " .09, ^J6
^25.66
^/^a>ud 3, /S55.
tJnecdene '^oi^/
(j.) If Mr. French had paid Mr. Gay $2 on the 1st of
February, and worked for him the 4. days ending Feb. 24
at $1.50 per day, the first part of the bill would Jmve been
made out as before, and then the credit entries would have
been made as follows : —
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180 BILLS OF GOODS.
M>^ /, ^ ''Sad, . . . . / S.OO
" S4, " 4t^^J/iJ<n,ai^/.50,^ 6.00
/ d.oo
^4. y«^ X 9}., . p7-66
xjneoaote ^^y*
( k.) The mere sending of a bill to a person, except at his request,
or at the time of sending the articles for which the biil is made out, ia
equivalent to a request that the money due on it should be paid.
(/.) A bill is said to be against the person who owes, and in favor of
the one who is to receive the money due on it. Thus the first of the
preceding bills is against Humphrey Barrett, and in favor of (Jeo. W.
Dodge.
100* Examples for Practice,— Due BiUs.
Make out the proper bills for each of the foUowing exam-
ples : — .
1. L. H. Holmes is a dry goods dealer, residing in Bridge-
water. Oct. 7, 1854, he sold to E. C. Hewett, for cash, 3
yds. of broadcloth at $3.62, 3 yds. of doeskin at $1.68, 1
cravat for $1.50, 1 vest for $6.00, 1 pair of gloves for $1.00.
and 1 pair of boots for $4.50.
2. A. B. Curry & Son, of Providence, sold to Geo. A,
Richards, June 13, 1855, the following articles, viz.: 85
* This entry is repeated so as to make the form of the bill more ap*
parent.
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BILLS OF GOODS. 131
rbs« live geese feathers, at 50 cents per pound ; 1 2 common
chairs, at $.42 each ; 6 cane seat, at $1.00 each ; 6 mahogany
spring seat, at $3.00 each ; 3 common bedsteads, at $3.00 each;
2 cottage bedsteads, at $5.00 each; 2 best hair mattresses, 25
lb. each, at $.50 per lb. ; 2 palm leaf mattresses at $4 each, 2
husk at $5.00 each, 2 straw at $2.50 each, 1 sofa, $35 ; 1
pair tete-a-tetes, $75 ; 1 gilt mirror, $75 ; 1 marble centre
table, $42 ; 1 secretary, $45 ; 1 painted chamber set, $45 ;
1 enamelled chamber set, $125.00 ; 3 common bureaus at
$8.00 each ; 1 marble toilet bureau, $42.00 ; 1 extension
dining table, $45.00 ; 12 oak dining chairs at $3.50 ; 1 time-
piece, $8.00 ; and 1 whatnot, $25.00.
3. John Smith sold to David Brown the following articles,
Tiz.: Oct. 7, 1854, 13 bushels of potatoes at $.56, 19 bu.
com at $.97, and 3 bbL of apples at $2.42; Nov. 1, 4 tons
of hay at $19, and 3 tons at $18.50 ; Dec. 17, 40 bushels
of potatoes at $.67, 34 bu. of corn at $.98, and 21 bbl. of
apples at $2.75 per barrel. Jan. 1, 1855, the account was
settled by a due bill.
Nqte. — A due bill is not a promise to pay a debt, but merely an
acknowledgment that it is due. It is intended to cat off after disputes
as to the debt for which it is given, bj famishing the creditor additional
means of establishing the jastice of his claim. Every dae bill, or writ-
ten promise to pay money, should contain the words "value received,"
to show that the person who signs it has received an equivalent for it.
Indeed, it is a principle in law, that no claim is valid unless it is based
on some service rendered, or consideration given^ to the person against
whom it is made, or on his account.
To illustrate the form of a due bill, we will suppose that John Smith
owes James Brown $25, and that he gives him a dije bill for the amount,
as follows : —
do^^a^^, ^t vtiuic received.
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132 BILLS OF GOODS.
The receipt, when a dae bill is given, might be, ** Received pajment
by due bill," or, " Settled by due bill."
4. May 7, 1855, Hill & Saunders, of Boston, sold to James
Drew, 1 cask linseed oil, 24 gal., at $1.50 per gal. ; 1 bag
Java coffee, 122 lbs., at 16 cents per lb. ; 1 hhd. of N. O.
molasses, 127 gals., at 34 cents per gaL ; 1 chest tea, 84 lb.,
at 48 cents per lb. ; 1 bag pepper, 24 lbs., at 11 cents per lb. ;
I box N. O. sugar, 278 lbs., at 6 cents per lb.
5. April 21, 1855, French & Simmons, of New York,
sold to Clark & Hubbard, 1 piece super, broadcloth, 26 yds.,
at $3.75 per yd. ; 1 piece fancy cassimere, 81 yds., at $1.34
per yd. ; 1 piece black cassimere, 23 yds., at $1.62 per yd. ;
6 pieces English prints, 29, 31, 30, 33, 29, and 31 yds., or
183 yds., at $.12 per yd. ; 3 pieces Merrimack prints, 28, 32,
31 yds., or 91 yds., at $.09 per yd. ; 2 pieces Scotch gingham,
42 and 41 yds., or 83 yds., at $.17 per yd. ; 1 piece sarsenet
cambric, 32 yds., at $.28 per yd. ; 3 dozen cotton hose at $1.87
per doz. ; and 2 lbs. Marshall's linen thread at $1.75 per lb.
6. Geo. Stevens, of Worcester, sold to Daniel Barnard,
Sept. 15, 1854, 28 bu. of potatoes at $.87 ; Oct 1, 43 bu.
of potatoes at $.65 ; Oct 7, 7 tons of hay at $19.50, 4 tons
at $18.37, and 2 tons at $17.25 ; and Nov. 3, 23 bbl. of
apples at $1.75, and 19 bbl. at $1.94. Mr. Barnard paid Mr.
Stevens, Oct 5, 1854, $20, Oct 17, $13.50 ; and Nov. 20,
he sold him 14 lb. sugar at 7 cents, 12 lb. at 9 cents, 8 lb. at
II cents, 7 lb. coffee at 15 cents, 4 lb. tea at 54 cents, 3 lb. at
47 cents, 4 lb. chocolate at 19 cents, and 1 bag salt for $1.25«
Jan. 1, 1855, Mr. Stevens made out his bilL
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PBOPVRTIES OF MUMBEBS, ScO. ISi
SECTION IX.
PROPERTIES OF NUMBERS, TESTS OF DIVISI.
BILITY, FACTORS, MULTIPLES, DIVISORS-
101. DefinitioM.
(a.) A FACTOR of anj given number is such a number as
taken an entire number of times will produce the given num-
ber ; or, the factors of a number are the numbers which
multiplied together will produce it.
Thus, 2 is a factor of 4, 6, 8, &c. ; 3 is a factor of 6, 9, 12, 15, Ac.
(h.) A DIVISOR of a number is any number which wiU
exactly divide it.
NoTB. — Every divisor of a nnmber must be a factor of it, and every
factor of a number a divisor of it The terms factor and diviaor, as
here used, are nzAy applied to entire numbers.
(c) A PRIME NUMBER is onc which has no other factors
besides itself and urntj.
Thus. I, 2, 3, 5, 7, 11, 13, 17, &C., are prime numbers.
(d.) A COMPOSITE NUMBER is One which has other fac-
tors besides itself and unity.
Thus, 4, 6, 8, 9, 10, 12, 14, 15, 16, &C., are composite numbers.
(«.) Any entire number of times a given number is a
MULTIPLE of it ; or, a multiple of a number is any number
which can be exactly divided by it
Thus, 12 is a multiple of 1, 2, 3, 4, 6, and 12, because it is an exact
nnmber of times each of them, or because it can be divided by each
without a remainder.
(/.) Two numbers are prime to each other when
they have no common factor.
For example, 4 and 9 are prime to each, as are 8 and 15, 24 and
35, &c
Again, 6 and 9 are not prime to each other, because they have the
common factor 3 ; 8 and 12 are not prime to each other, because they
have the common factor 4, &c
12
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lo4 PROPERTIES OF NUMBERS, &C.
NoTB. — It is obTious from the foregoing:, that every number (■ ft
fiictor of all its multiples, and a multiple of all its factors.
103* Demonstration of Principles.
Proposition Firsf^ — If one of two numbers is a factor of
another^ it must he a factor of any number of times that other
number.
For to find any number of times a given number, we have
only to multiply the number by some new factor, without
striking out any of the former ones.
Illustrations. — Since 2 is a factor of 12, it must be a factor of any
number of Umes 12, as 24, 36, 48, &c
Since 7 is a factor of 14, it must be a factor of any number of
times 14, as 28, 42, 56, &c.
Proposition Second. — If each of two numbers is a multi'
pie of a third number^ their sum and their difference must
also be multiples of that third number.
For, adding an exact number of times a given number to,
or subtracting it from, an exact number of times the same
number, must give an exact number of times that number.
Illustrations. 8 times 9, or 72, -f- 3 times 9, or 27, = 11 times 9, or
99. So 8 times 9, or 72, — 3 times 9, or 27, = 5 times 9, or 45.
Again. Both 12 and 20 are multiples of 4, and so is their snm, 32,
and their difference, 8.
Both 42 and 28 are multiples of 7, and so is their sum, 70, and their
difierence, U.
Proposition Third, — If one of two numbers is a multiple
of a third number, and the other is not, neither their sum nor
their difference will be a multiple of that third number.
For both the sum and the difference of an entire, and a
fractional, number of times a given number, must equal a
fractional number of times that given number.
Illustrations, 8 times 6, or 48, -f- 2± times 6, or 15, = 101 times 6,
or 63.
8 times 6, or 48, — 21 times 6, or 1 5, = 51 times 6, or 33.
71 times 9, or 66, - - 4 times 9, or 36, = 31 times 9, or 30.
7^ times 9, or 66, + 4 times 9, or 36, = 11^ times 9, or 102.
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PROPERTIES OF NUMBERS, &C. 185
A^in. 20 is a mnltiple of 5, nnd 13 is not; hence, ncithct their
sum, 33, nor their difference, 7, is a multiple of it.
38 is not a multiple of 6, and 24 is ; hence, neither their sum, 62,
nor their difference, 14, is a multiple of it.
Proposition Fourth. — Jf neither of ttao numbers is a mill'
tiple of a third, their sum or their differeTice may or may not
he a multiple of it
The truth of this proposition can best be made manifest by
a few illustrations.
1 . Neither 7 nor 23 is a multiple of 2 ; jet hoth their snm, 30, and
their difference, 16, are multiples of 2.
2. Neither 5 nor 28 is a multiple of 3; jet their snm, 33, Is, and
their difference, 23, is not, a multiple of 3.
3. Neither 8 nor 17 is a multiple of 3 j jet their sum, 25, is not, and
their difference, 9, is, a multiple of 3.
4. Neither 14 nor 27 is a multiple of 4 ; and neither their sam, 41,
nor their difference, 13, is a multiple of 4.
103. Tests of the Divisibility of Numbers.
Application of the foregoing propositions.
1. Dimsibility by 2, 5, 8^, or by any other number which
will exactly divide 10.
Every number greater than ten is composed of a certain
number of tens, plus the units expressed by its right hand
figure. But the part which is made up of tens must (lOS,
Prop. I.) be divisible by any divisor of ten ; and hence,
(10^9 Prop. II. and III.) the divisibility of the entire num-
ber will depend on the part expressed by the right hand
figure.
Therefore, a number is divisible 5y 2, 5, 2J, 1§, or 5y any
other number which will exactly divide 10, when its right hand
figure is thus divisiUe.
Illxmtration. 4125 is divisible by 5, bj 2^ and bj 1^ because each
of these numbers will exactlj divide 10, and also 5, the right hand figure
of the given number.
n. Divisibility by 4, 20, 25, 50, 12 J, 16§, or any other
number which will exactly divide 100.
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186 PROPERTIES OF KITMBKBS, &C.
Every number greater than 100 is composed of a certaiik
number of hundreds, plus the number expressed by its two
right hand figures. But the part which is made up of hun-
dreds must (103) Prop. I.) be divisible by any divisor of
one hundred, and hence, (I03, Prop. U. and III.) the divisi-
bility of the entire number must depend on the part expressed
by the two right hand figures.
Therefore J a number is divUihU hy 4, 20, 25, 50, 12^, or
hy any other number which wiU exactly divide 100, where
its two right hand figures are thus divisible,
III. Divisibility by 8, 40, 125, 250, 500, 333 J, 166|, or hy
any other number which vnU exactly divide 1000.
Every number greater than 1000 is composed of a certain
number of thousands, plus the number expressed by its three
right hand figures. But the part which is composed of thou-
sands must (103^ Prop. I.) be divisible by any divisor of
1000, and hence, (!©», Prop. II. and III.) the divisibility
of the entire number must depend on the three right hand
figures.
Therefore a number is divisible by 8, 125, 250, 500, 166§,
833^, or by any number which will exactly divide 1000, when
its three right hand figures are thus divisible.
Note. — Similar tests for determining the divisibility of numbers by
any divisor of 10,000, 100,000, &c., could be established, but as they
could very rarely be applied to advantage, we omit them.
IV. Divisibility by 9.
(a.) If 1 be subtracted from a unit of any decimal denom-
ination above unity, the remainder will be expressed entirely
by 9*s, and will therefore be a multiple of 9.
Illustrations. —
10—1=9
100 — 1 = 99
1000 — 1 = 999
lOOOO — 1 = 9999
&c., &c.
(5.) But unity, or 1, = X 9 -f- 1 ; and if, for convenience
^ statement, we regard X 9, or 0, as a multiple of 9, it
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PROPERTIES OF NUSIBERS, &C. 137
will follow that a unit of anj decimal denomination is 1 more
than a multiple of 9.
Illustrations. —
1 =0 + 1
10 = 9 + 1
100 = 99 + 1
1000 = 999+1
&c., &c.
(c.) As a unit of any decimal denomination is 1 more than
a multiple of 9^ 2 units must be 2 more than such a multiple,
8 units must be 3 more, 4 units 4 more, 5 units 5 more, &c
Illustrations. — Since 1000 = 1 more than a multiple of 9, 7000 must
equal 7 more.
Since 1000000 = 1 more than a multiple of 9, 7000000 must equal
7 more, &c.
{d.) But the digit figures of a number express the number
of units of its various denominations, and therefore any num«
ber must be as many more than a multiple of 9 as there are
units in the sum of its digit figures.
lUustrations. 8235 = 8000 + 200 + 30 + 5.
8000 = 8 more than a multiple of 9.
200 = 2 more than a multiple of 9.
SO = 3 more than a multiple of 9.'
5 = 5 more thun a multiple of 9.
Therefore, 8235 = 8 + 2 + 3 + 5. or 18 more than a multiple of 9,
or it equals a multiple of 9, plus 18, and must therefore (since 9 is a
divisor of 18) be a multiple of 9.
Again. 57864 = 50000 + 7000 + 800 + 60 + 4.
60000 = 5 more than a multiple of 9.
7000 = 7 " " " " " "
800 = 8 " " " « ** "
60 = 6 " " " " " "
A ^S A U U U tC C( t(
Therefore, 57864 = 5 + 7 + 8 + 6 + 4, or 30 more than a multi-
ple of 9, and is therefore (since 9 is not a divisor of 30) not a multiple
of 9.
(«.) From these principles it follows, —
1. That every nttmher is equal to a multiple of 9, plus ih^
sum of its digit figures.
12*
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138 PROPERTIES OF NUMBERS, &C.
V
2. That if the mm of tlie digit fgures of any number bt
subtracted fro7n it, the remainder will be a multiple of 9.
3. That a number is divisible by 9 when the sum of %t$
digit figures is thus divisible,
4. That the remainder obtained by dividing the sum of the
digit figures of any number by 9, is the same as that obtained
by dividing the number itself by 9.
5. Thai the difference of any two numbers, the sums of
whose digits are alike, wiU be a multiple of 9.
6. That the divisibility of a number by 9 wiU not be affect"
ed by any change in the order of its digits.
7. That if the digit figures of a number be added together,
and then the digit figures of the result, and so on till the sum
is expressed by a single figure, that figure will either be 9, or
the remainder obtained by dividing the original number by 9.
^ the figure is d, the number is a mtdtiple of 9.
Eemark. — Finding the excess of any number over a multiple of 9
is called casting out the 9's.
(f.) Consequences of the foregoing, — From the foregoing properties
«f the number 9, considered in connection with the principles established
in 1 03^ come some convenient methods of proving numerical opera-
tions ; a few of which we will mention, leaving the pupil to find out the
reasons for each.
1. To prove Addition. — Cast out the 9*8 from the several numbers
added, add the results, and cast out the 9*s from their sum. Then cast
out the 9's from the number obtained as the answer to the question, and
if the work be correct, the last two results will be equal.
2. To prove Subtraction. — Cast out the 9's from the subtrahend and
remainder, add the results, and cast out the 9's from their sum. Then
cast out the 9*s from the minuend, and if the work is correct, the last
two results will be equal.
3. To prove Multiplication. — Cast out the 9*s from the several factors
employed, multiply the results together, and cast out the 9*s from
their product. Then cast out the 9*8 from the product of the original
multiplication, and if the work is correct, the last two results will be
equal.
4. To prove Division. — Cast out the 9*s from the divisor, quotient,
ftnd remainder; to the product of the first two results add the last
result, and cast out the 9*8. Then cast out the 9*8 from the dividend,
and if the work is con-ect, the last two results will be equal
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PROPERTIES OF NUMBERS, &C. 139
ig.) Dependent on the same principles is the following, nhich th«
papil may use with the xminitiated as an arithmetical puzzle.
To tell what figure has been erased. — Tell a person to write any nnnk
ber whatever, without informing you what it is ; to subtract the sum of
its digit figures from it ; to erase from this result any digit figure, other
than zero, and write zero in its place ; and finally, to add together the
digit figures of the number thus obtained, and tell you their sum.
The difference between this sum and the next higher multiple of 9
will show the figure removed. Thus, if the sum is 29, 7 was erased ; if
it is 45, 9 was erased ; &c.
Let the pupil explain the reasons of this.
V. DimsihiKty hy 3.
Since 9 is a multiple of 3, eveiy number which is a multi-
ple of 9 must also be a multiple of 3, (Prop. I.) Therefore,
a unit of any decimal denomination must be 1 more than a
multiple of 3 ; and hence a number i$ a mvUiple of 3 when
the sum of its digit figures ts such a multiple,
VI. Divisibility by Ih
(a.) A unit of any decimal denomination is either 1 or 10
more than a multiple of 11. Thus, —
1=0X 11 + 1
10 = X 11 + 10
100 = 9 X 11 + 1
1000 = 90 X 11 + 10
10000 = 909 X 11 + 1
100000 = 9090 X 11 + 10
Or, arranging the numbers with reference to the remainden, w«
have —
1=0 X 11 + 1
100 = 9 X 11 + 1
10000 = 909 X 11 + 1
1000000 = 90909 X 11 + 1
10 = 0X 11 + 10=1 X 11— 1
1000 = 90 X 11 + 10 = 91 X 11 — 1
100000 = 9090 X 11 + 10 = 9091 X 11 — 1
10000000 = 909090 X 1 1 + 10 = 909091 X 11 — 1
(b,) From which we see that a unit of any denomination
expressed by a figure occupying the 1st, 3d, 5th, or any other
odd place from the point, is 1 more than a multiple of II ;
and that a unit of any denomination expressed by a fignr*
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140 PROPERTIES OP NUMBERS, &C.
occupying the 2d, 4th, or any other even place from tht
point, is 1 less than a multiple of 11.
(c.) Hence, on account of the figures occupying odd places
from the point, a numher is as many more than a multiple of
11 as there are units in the sum of these figures, while on
account of the figures occupying even places, it is as many
less than a multiple of 11 as there are units in the sum of
those figures.
(rfL) Hence, every number is equal to some multiple of 11,
plus the sum of its digit figures occupying odd places from
the point, minus the sum of those occupying even places. If
these sums are alike, the additions will equal the subtractions,
and the number will be a multiple of 11. If these sums are
unlike, their difference will be the excess of the additions
over the subtractions, or of the subtractions over the ad-
ditions.
(c.) If, then, the difference of the sums of the alternate
digits is a multiple of 11, the whole number will be either
the sum or the difference of two multiples of 11, and hence a
multiple of 11 ; but if this difference is not such a multiple,
the whole number will be either the sum or the difference of
two numbers, one of which is, and the other is not, a multiple
of 11, and hence will not be a multiple of 11.
(yi) Jlence, a number is a multiple of 11, when the sums
of its alternate digits are equal, or when their difference is a
multiple of 11.
First Example. — Is 15873 a multiple of 11 ?
Solution. — The sum of the digits in the odd places is 3 + 8 + ^ or
12 • the sum of those in the even places is 7 -h ^) or 12. Hence, the
two sums are alike, and 15873 is a multiple of 11.
Second Example, — Is 274854 a multiple of 11 ?
Solution. — The sum of the digits in the odd places is 4 -f- 8 + 7 =»
19 ; the sum of the digits in the even places is 5 -|"^ + ^ ^^ 11 i the
dl&rence between 19 and 11 is 8, which is not a multiple of 11. Hcnoe^
9748M U not a multiple of 11.
VJi# J^ a number is divisible by each of two nwrJ^en
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PROPERTIES OF NUMBERS, ScC, 141
which are prime to each other, it toiU be divisible h^ their
product.
For dividing by one cannot (since the numbers are prime
to each other) cast out the other, or any factor of it.
Illustrations* I. A number which is divisible by 4 and 9 mutt bd
dirisible by 4 X 9, or 36.
2. A number which is divisible by 8 and 15 must be divisible by 8 X
15, or 120
3. A number is divisible by 12, when it is by 3 and 4.
4. A number is divisible by 35, when it is by 7 and 5.
5. A number is divisible by 42, when it is by 6 and 7.
Vm. If a number is divisible by each of two numben
which have a common factor, it will not of necessity be divisi*
ble by their product.
For dividing by one must cast out the common factor of
the two numbers, and if that factor be not taken more than
once as a factor of the original number, the quotient will not
be divisible by the other of the two numbers.
Illustrations. 84 is divisible by 4 and 6, but not by their product, 24.
72 IS divisible by 4 and 6, and also by their product, 24.
IX. j^ one number is not divisible by another, it vnU not
he divisible by any multiple of that other number.
For if a number does not contain once another number, it
cannot contain any number of times that other number.
Illustrations. — A number which is not divisible by 2. is not divisible
by 4, 6, 8, 10, &c A number which is not divisible by 3, is not divisi-
ble by 6, 9, 12, 15, 18, 21, &c.
X. A number which is divisible by any composite number
is divisible by all the factors of that composite number.
For dividing by any composite number is merely dividing
by the product of its factors.
104:* Recapitulation, for convenience of reference,
I. Any number is divisible by 2, 5, 3^ or any other number which wHH
exactly divide 10, when its right hand figure is thus divisible.
II. Any number is divisible by 4, 20, 25, 50, 12^, 16|, or any olher
number which will exactly divide 100, when the number expressed by its tw0
right hand figures is thus divisible.
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142 PROPERTIES OP NUMBERS, 5cO.
IIL An^ number is divisibU by 8, 4, 125, 250, 333 1, or any other mtm
her which will exactly divide 1000, when the number expressed by its threi
right hand figures is thus divisible.
TV. Any number is divisible by 3 or 9, witen the sum of its digits is thus
divisible.
y. Any number is divisible by 11^ when the sums of its alternate digiU
are equal, or their difference is a multiple o/* 11.
VL A number which is divisible by each of two numbers which are primf
to each other, is divisible by their product.
Vn. A number which is divisible by each of two numbers not prime ta
each other, is not of necessity divisible by tlieir product.
YIIL A number which is not divisible by another is not divisible by any
multiple of that other number.
IX. A number which is divis3>le by any composite number is cdso divisir
hie by all the factors of that composite Clumber.
105* Definitions of Factors, Powers, S^c.
(a.) A number is said to be divided into factors when any
factors which will produce it are found.
Thas, in 36 = 4 X 9, 36 is divided into the factors 4 and 9 ; but in
86 a= 2 X 6 X 3 it is divided into the factors 2, 6, and 3.
(h.) A number is said to be divided into its prime factors
when it is divided into factors which are all prime numbers.
Illustrations. 36 = 2X2X3X3 8 = 2X2X2
30 = 2X3X5 84 = 2X2X3X7
(c.) When any number is taken more than once as a factor
to produce another number, we may express the number of
times it is taken as a factor, by placing a small figure above
it and a little to the right.
Illustrations. 3^ means the same as 3 X 3 ; i. e., that 3 is to be taken
twice as a factor.
8* means the same as 3 X 3 X 3 X 3 ; i. e., that 3 is to bo taken 4
times as a factor.
38 X 2< means the same as 3 X 3 X 3 X 2 X 2 X 2 X 2; i.e., that
the product of 3 taken 3 times as a factor is to be multiplied by tlio
product of 2 taken 4 times as a factor.
Note. — The student should notice the difference between taking a
number as a factor a rertain number of times, and merely taking it ft
number of times.
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METHOD OP FACTORING NUMBERS. 148
Thus, 5 <alcen 3 times as a factor =s 5 times 5 times 5 =» 125, bat 5
kikm 3 times = 3 times 5 = 15.
(d.) The product of a number taken any number of times
as a factor is called a power of the number.
Illustrations. 9 is the second power of 3, becanse it is the product of
3^« i. e., of 3 taken tnrice as a factor.
2^ is the fifth power of 2, because it is the product of 2*, i e., of 2
taken 5 times as a factor.
(e.) The figure indicating how many times a number is
taken as a factor is called the exponent of the power to
which the number is raised.
Thas, in 2^, the exponent is 5 ; in 5% it is 2 ; in 6^, it is 4 ; &c.
(/) The following examples indicate the method of r-sad-
ing numbers expressing powers.
3^ is read three seventh power^ or three to the seventh power.
4* X 2^ is read four fifth power multiplied by two third power.
Note. — The second power of a number is sometimes called itf
SQUABS, and the third power its cube.
106« Method of Fcu:toring Numbers.
1. What are the prime factors of 2772 ?
Solution. -^Wq see by 104y 11., that 2772 is divisible by 4. and
therefore by 2 X 2, the prime factors of 4.
Dividing by 4 gives 693 for a quotient, which, by 104^ IV^ we see
is divisible by 9, and therefore by 3 X 3, the prime factors of 9.
Dividing 693 by 9 gives 77 for a quotient, the prime factors of which
are 7 and 11. Hence, the prime factors of 2772 are 2>, 3*, 7, and 11 ;
or, which is the same thing, 2772 = 23 X 32 X 7 X 11.
2. What are the prime factors of 29766 ?
Solution. — We see by 104, 1, and IV., that 29766 is divisible by
both 2 and 3, and hence by their product, 6. Dividing by 6 gives 4961
for a quotient, which, by 104^ V., is a multiple of 11. Dividinjr by 11
gives 451 for a quotient, which (I.O49 Y.) is also a multiple of 11
Dividing by 11 gives 41 for a quotient, which is obviously a prime
number. Therefore, 29766 = 2 X 3 X ll^ X 41.
3. What are the prime factors of 3871123 ?
&Aution. — We readily see that it is not divisible by 2, 3, fi, or II,
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144 FACTORS OF NUMBERS.
and therefore we must see whether it is divisible by the oAer prime
numbers, beginning, for convenience sake, with the smallest Bj actual
trial, we find that 7, 13, 17, 19, and 23 each leave a remainder after
division, bat that 29 is contained in it exactly 133487 times. Therefore,
29 is one of the prime factors of the given number, and the others will
be fband in 133487. Bat no nnmber less than 29 can be a factor of
133487, for none is a factor of the original nnmber. We therefore first
try 29, which we find is contained exactly 4603 times. Therefore, 29 is
again a factor of the original number, and the remaining factors must
be found in 4603. But no number less than 29 can be a factor of 4603,
for none was a factor of the original number. Beginning with 29, we
try in succession 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, and 71, and find
that none of them will exactly divide 4603 ; and we observe, moreover,
that the entire part of the quotient of the division by 71 is less than 71.
Now, as the divisor increases, the quotient must decrease ; therefore,
if there is any number larger than 71 which will divide it, the quotient
most be less than 71. But when a division can exactly be performed,
the divisor and quotient must each be a factor of the dividend ; and
hence if 4603 has a factor greater than 71, it must also have one less
than 71. But as we have already found that it has no factor less than
71, we infer that it can have none greater, and that it is a prime number.
Hence, 3871123 =: 29^ X 4603.
(a.) The labor of testing the divisibility of a number by
the yarious prime numbers may be made much less laborious
and tedious, by considering what must be the last figure of
the quotient, if the division by any number is possible.
(6.) Thus, in the last example, in testing the divisibility of 4603, we
may observe that if 29 is a factor of it, the last figure of the other factor
must be 7, for 7, or some number ending in 7, is the only number which,
multiplied by 29, will give for a product a number ending in 3. Divid-
ing by 29, we have, 29 is contained in 46 once, and 17 remainder; in
170, 5 times, and 25 remainder; and we can readily see that it is con
tained in 253 more than 7 times.
(e.) Again. If 31 is a factor of 4603, the last figure of the other
factor must be 3, for 3, or some number ending in 3, is the only number
which, multiplied by 31, vrtll give for a product a number ending in 3.
Dividing by 31, we have, 31 is contained in 46 once, and 15 remainder;
in 150, 4 times, and 26 remainder ; and we can see at a glance that it is
contained in 263 more than 3 times.
(d.) Again. If 37 is a factor of 4603, the last figure of the other
factor must be 9. But 37 is contained in 46 once, and 9 remainder ; in
90, twice, and 16 remainder ; and it is obvious that it cannot be con
tained as ma&y as 9 times in 163.
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FACT.>RS OF NUMDERl.
145
(e,) From the above, we see that to find the prime factors
*if a number, we may first divide it by any number which will
ilivide it without a remainder, then divide this quotient by any
number which will divide it without a remainder, and so pro-
ceed with each successive quotient, tiU wc reach one which is
H piime number, or which can be readily divided into its
prime factors. The prime factors of the several divisors and
of the last quotient will be the prime factors required.
(/.) If in any case we cannot discover a divisor of any
given number by the tests of 104, we try in succession 7,
and all the prime numbers from 13 upwards, till we find one
which is a divisor of the given number, or till the entire part
of our quotient is less than the number employed as a divisor,
in which case the number is prime.
107. Exercises for the Student,
(a.) Find the prime factors of the numbers from 1 to 100,
writing them as in the following model.
1, prime.
7, prime.
2, prime.
8 = 2X2X2 = 2«.
3, prime.
9 = 3 X 3 = 32.
4 = 2 X 2 = 2«.
10 = 2 X 5.
5, prime.
11, prime.
6 = 2X3.
12 = 2 X 2 X 3 = 29 X 3
(h,) We would recommend that the pupil make out a table
of the prime factors of the numbers from 1 to 1000. He
will find it a very profitable exercise, and one which will
greatly aid him in all his subsequent work.
(c.) What are the prime factors —
1.
Of 1001 ?
8.
Of 1183?
2.
Of 1025 ?
9.
Of 1625?
8.
Of 1024?
10.
Of 2057 ?
4.
Of 1033 ?
11.
Of 16128?
5.
Of 1096 ?
12.
Of 3809 ?
6.
Of 1157?
13.
Of 6381 ?
7
Of 1067 ?
IS
14.
Of 7128?
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146 Divisoua OF mimhers.
15.
Of 7854 ?
21.
Of 444528 ?
16.
Of 5989 ?
22.
Of 1072181 ?
;7.
Of 5625?
23.
Of 5764801 ?
18.
Of 9257 ?
24.
Of 6103515625?
19.
Of 10917?
25.
Of 12168587041J ?
20.
Of 843479 ?
108* Common Divisor^ DefinitionSy and Properties,
(a.) A DIVISOR of a number is a number which will ex-
actly divide it.
(b,) A COMMON DIVISOR OP TWO OR MORE NUMBERS
is a number which is a divisor of each of them.
(c.) The GREATEST COMMON DIVISOR OP TWO OR
MORE NUMBERS is the largest number which is a divisor of
each of them.
{cL) From these definitions and the principles previously
established, it follows that, —
1. A divisor of a number can contain only stick prime fac-
tors as are found in that number,
2. A common divisor of two or more numbers can contain
only such prime factors as are common to all the numbers.
3. The greatest common divisor of two or more numbers is
the product of aU the prime factors common to all the given
numbers.
109* Greatest Common Divisor. — Method by Factors.
1. What is the greatest common divisor of 819 and 1071 ?
Solution, — The greatest common divisor of 819 and 1071 is the
product of all the prime factors common to those numbers. <\
819 = 32 X 7 X 13
1071 = 32 X 7 X 17
From which we see that the only common factors are 3* and 7. There*
fore, 32 X 7, or 63, is the greatest common divisor required.
What is the greatest common divisor —
2. Of 792 and 936 ? I 3. Of 1125 and 1575 ?
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niVISORS OF NUMBERS. 147
4. Of 3102and3Gr)6?
5. Of 287 and 369 ?
6. Of 4652 and 5544?
7. Of 924 and 1188?
8. Wtjit is the greatest common divisor of 72, 108, and
252 ?
aSoIiUioii. — The greatest common divisor of 72, 108, and 253, is the
pr^xluct of all the prime factors common to thor^ numbers.
72 = 28 X 32
108 = 22 X 33
252 = 22 X 3« X 7
Fiom which we see that the only common factors are 2* and 3*. Thero>
fore, 32 X 22, or 36, is the greatest common divisor required.
"What is the greatest common divisor —
9. Of 168, 505, and 539?
10. Of 1386, 3234, and 4158 ?
11. Of 864, 2058, and 2346 ?
12. Of 686, 1029, and 2401 ?
13. Of 112, 147, 168, and 189 ?
14. Of 576, 672, 864, and 1132 ?
IIO. A more brief Method,
(a.) The following solution will usually be found much
more brief than the preceding, inasmuch as it avoids the
necessity of separating all the numbers into their factors. By
it, we find the factors of any one of the numbers, and then see
which of them ai'e factors of all the other numbers.
1. What is the greatest common divisor of 756, 840, 1386,
and 1596 ?
Solution. — We first find the factors of 840, because they can be the
most readily found. 840 = 2^ X 3 X 5 X 7, and we have now to find
which of these factors, if any, are factors of all the other given num»
bers. It is obvious (104, I.) that 2 is a factor of all of them, and
(104, II.) that it is contained but once as a factor in 13S6. Hence, 2
will enter once only as a factor of the greatest common divisor. 5 is
(104y I.) a factor of no other number. 3 being a factor (104, IV.)
of all the numbers, is a factor of the greatest common divisor. By
trial, 7 is found to be a factor of all the numbers, and hence of tlie great-
est common divisor. Hence, tJ e greatest common divisor is 2 X 3 X
7. or 42.
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148 DIVISORS OF NI;M55KR3.
(b,) What is the greatest common divisor —
2. Of 3465, 4375, and 5^50 ?
3. Of 1792, 936, 1224, and 1056?
4. Of 1342, 1738, 237(. and 250C?
6. Of 3312, 6048, 957G, and 6330?
6. Of 4572, 2380, 5272, and 8304 ?
7. Of 3125, 4379, 8243, and 5975 ?
111. Factoring not always necessary,
(a.) We can frequently find the greatest common divisor
of the given numbers without finding their prime factors.
(6.) Thas, in finding the greatest common diyisor of 24 and 42, wo
can see at a glance that 6 is a divisor of both ; that 24 s=: 6 X 4, and
42 s= 6 X 7 ; therefore, since 4 and 7 have no common factor, 6 mast
be the greatest common divisor required.
(c.) Again. In finding the greatest common divisor of 693 and 819
we see at once that 9 will divide each.
693 = 9 X 77
819 = 9 X 91
Now, comparing 77 and 91, we see that 7 will divide each.
77 = 7 X 11
91 = 7 X 13
And as 7 and 11 are prime to each other, there is no other common fac>
tor to the given numbers. Hence, 7 X 9, or 63, is the greatest common
divisor required.
(d.) Again. In finding the greatest common divisor of 36 and 72,
we may see that 36 is a divisor of 72, and hence the greatest ouhimoii
divisor required.
(c.) What is the greatest common divisor —
1. Of 12 and 18 ?
2. Of 28 and 42 ?
3. Of 18 and 54 ?
4. Of 48 and 84?
5. Of 324 and 594?
6. Of 2025 and 2916?
7. Of 5544 and 11583?
8. Of 18, 48,72, and 66?
9. Of 12,36, 60, and 132?
10. Of 49, 63, 84, and 91 ?
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DIVISORS OF NUMBERS. 149
1 13. Method hy Addition and Subtraction.
(a.) When the sum or the difference of any two of the
given numbers is a small number, or one more easily divided
into factors than any of the original numbers, the work may
be abbreviated by applying the principles of IOS9 Propo-
sition II.
1. What is the G. C. D * of 8379 and 8484 ?
Suggestion. — The G. C. D. required must be a divisor of 8484 —
8379, which i8l05 = 3X 5X 7; and we have only to ascertain which
of these factors, if any, arc factors of one of the given nambers.
2. What is the G. C. D. of 89437, 95429, and 90537 ?
Suggestion. — The G. C. D. required must be a divisor of the differ-
ence of any two of the given numbers, as of 89437 and 90537, which is
1100 = 2^ X 52 X 11 ; and we have only to ascertain which of these
factors, if any, are factors of all the given numbers.
3. What is the G. C. D. of 56474 and 28526 ?
Suggestion. — The G. C. D. required must be a divisor of 56474 -f-
28526, which is 85000 = 2^ X 9^* X 17 ; and we have only to ascertaiq
which of these factors, if any, are factors of one of the given numbers.
4. What is the G. C. D. of 3598, 5383, 6545, and 8617 ?
Suggestion. — The G. C D. required must be a divisor of the sum of
any two of the given numbers, as of 5383, and 8617, which is 14000 =»
2* X 53 X 7 ; and we have only to ascertain which of these factors, if
any, are factors of all the given numbers.
What is the greatest common divisor —
5. Of 949 and 962 ?
6. Of 857637 and 857692 ?
7. Of 5489 and 7689 ?
8. Of 9709, 10906, and 10241 ?
9. Of 71227, 72553, 73840, and 75127 ?
10. Of 930069, 992673, and 1103673?
11. Of 12551 and 25949 ?
12. Of 169881 and 170119?
13. Of 16181 and 16324?
14. Of 180006, 293694, 468963, and 596862 ?
♦ G. C. D. means greatest common divmr*
13'
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150 DIVISOUS OF NUMBERS.
113. Demonstration of Method hy Division.
(a.) Proposition. — The G. G, D, of any two numbers it
the same as the G. C. D. of the smaller, and the remaindef
left after dividing the larger hy the smaller,
(b.) To prove this, let the letter a represent any nnmber whatevei;
and the letter 6 represent any other number larger than a. Now, what-
ever numbers a and 6 represent, it is obvious that a is contained in h
some number of times, which we will call c times, and that there may,
or may not, be a remainder. If there is no remainder, a will be the
G. C. D. of a and 6. If, however, there is a remainder, we will call it d,
and we have to prove that the G. 0. D. of a and 6 is the same number
as the G. C. D. of a and d.
(c.) Heprescnting the division by writing the letters as we should the
numbers they represent, we have, —
Dividend.
Divisor a ) h ( c = Quotient.
aX c = Product of divisor by quotient
d = Kemainder.
(d.) From the nature of division, it follows that ^
d = b — a X c
and 6 = c? -j- « X c
(e.) Now, as any divisor of a must (10^9 Prop. I.) be a divisor of
c X Oj the G. C. D. sought must be a divisor of 6 and c X a, and henco
(105^, Prop. 11.) of 6 — c X a,OT d. Again. Any divisor of d and a
must be a divisor of d and c X a, and hence (lOd^ Prop. II.) of d +
f X a, or b. It therefore follows that the G. C. D. sought is the same
number as the G. C. D. of c? and a, which, as a and b may represent any
numbers whatever, establishes the proposition.
11 4. Application of the foregoing Principle.
(a.) When the numbers of which the G. C. D. are re-
quired are such as cannot easily be divided into factors, or
iolved by the methods heretofore given, the principle just
demonstrated may be advantageously applied, thus : —
(L) Divide the greater of any two numbers by the less ;
then will the remainder obtained by this division and the
smaller of the two numbers haye the same G, C, D, ^ the
'imbers themselves.
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DIVISORS OP NUMBERS. 151
(c ) But as the remainder after any division must always
be less than the divisor, we may divide the original divisor,
I. e., the smaller number, by this remainder. Now, if there
be a remainder from this division, it follows, from the propo-
sition, that the G. C..D. sought must be the G. C. D. of this
remainder and the last divisor ; and we may divide the last
divisor by the last remainder.
(d.) But as the remainders must constantly be decreasing,
it follows that if we continue this process, we shall at last find
a remainder which will exactly divide the preceding, and will
therefore be the G. C. D., or we shall have a remainder of 1,
m which case the numbers are prime to each other.
1. What is the G. C. D. of 4277 and 9737 ?
WRITTEN WORK.
4277 ) 9737 ( 2
1183 ) 4277 ( 3
728 ) 1183 ( 1
455 ) 728 ( 1
273 ) 455 ( 1
182 ) 273 ( 1
aT) 182 ( 2
00
Explanation. — Dividing the greater number by the less gives 1183
for a remainder. Therefore, the G. C. D. required is the G. C. D. of
the smaller number, 4277 and 1183. Dividing 4277 by 1183 gives 728
for a remainder. Therefore, the G. C. D. required is the G. C. D. of
728 and 1183. Dividing 1183 by 728 gives 455 for a remainder. There-
fore, the G. C. D. required is the G. C. D. of 728 and 1183.
l*rocceding in this way we find that the G. C. D. required is the same
as the G. C. D. of 91 and 182, which is 91.
(e.) The work can frequently be much shortened by ob-
serving that if any remainder contains a prime factor which is
not a factor of the preceding divisor, the factor may be cast
out without affecting the G. C. D.
Thus : 728, the second remainder in the above example, is obviously
a multiple of 8, or 2*"*, and 2 is not a factor of 1183, the preceding divi*
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152 MULTIPLES OF NUMBERS.
sor. We may therefore cast out the factor 8 from 728. The ottici
factor is 91 ; therefore, the G. C. D. required is the same as the G. C. D.^
of 91 and 1183, which is 91.
The work, then, would be written thus : —
4277 ) 9737 (2
1183 ) 4277 ( 3
728 = 8 X 91 ) 1183 ( 13
273
"oo
Hence, 91 = G. C. D.
(/.) What 18 the greatest common divisor of -
2. 8383 and 9589 ?
3. 2021 and 6493 ?
4. 6493 and 8909 ?
5. 425743 and 584159 ?
6. 113423 and 836987 ?
7. 6941849 and 9128111?
lltS. Common Multiples, Definitions, and Properties.
(a.) A multiple of a number is any entire number which
can be exactly divided by it.
(6.) A common multiple of two or more numbers is a num-
ber which is a multiple of all of them.
(c.) The least common multiple of two or more numbers
is the smallest number which is a multiple of all of them.
(cf.) From these definitions and the principles previously
established, it follows that, —
1. A multiple of a number must contain all the prime Jani-
tors of that number.
2. A common multiple of two or more numbers must con-
tain all the prime factors of each of them,
r 3. The least common multiple of two or more numbers must
be the smallest number which contains all the prime factots of
each of them,
116. Least Common Multiple, — Method by Factors,
1. What is the L. C. M.* of 504, 756, 924, and 1176 ?
* L. C. M. means least cotyimo7i muUiple.
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MULTIPLES OF NUMBERS. 158
Sciutton. — The L. C. M. of these numbers is the smallest nambef
' frhich contains ail the prime factors of each of them.
504 = 23 X 32 X 7
756 = 22 X 33 X 7
924 = 22X 3 X 7 X II
1176 = 28 X 3 X 72
The factors of 1 1 76 are 28 X 3 X 72 all of which we take. The fac
tors of 924 arc 22 X 3 X 7 X 11, all of which, except 11, we have
taken. We therefore introduce the factor 11, which gives 28 X 3 X 7«
X 11. The factors of 756 are 22 X 33 X 7, all of which, except 32, we
ha ve taken. We therefore introduce the factor 32, which gives 28 X 38
X 72 X 11. The factors of 504 are 23 X 32 X 7, all of which we have.
Therefore, the product of 23 X 3? X 72 X 11, which is 116424, is the
L. C. M. required.
Note. — When the factors of the L. C. M. have been obtained, much
labor in multiplying may be saved, by writing some one of the numbers
in place of its factors. Thus, in the example above, by writing 1176
Instead of its factors, we should have 1176 X 32 X 11 = 116424 =a
L. C. M., as before.
What is the least common multiple —
2. Of 18, 24, and 42 ?
3. Of 36, 48, and 60 ?
4. Of 132, 144, and 160?
5. Of 98, 126, and 186 ?
6. Of 364, 637, and 1547 ?
7. Of 605, 325, 715, and 1001 ?
8. Of 504, 756, 1008, and 1512 ?
9. Of 594, 1386, and 1782 ?
10. Of 735, 945, 1365, and 2310 ?
11. Of 154, 231, 264, and 392 ?
117. Abbreviated Method,
(a,) When the factors of the several numbers can readily
be determined, or after they are found, much labor may often
be avoided by considering what factors are wanted with any
one of ihe numbers to produce the L. C. M.
(6.) Thus, in the example solved in 116, since the factors cannot
easily be recognized, we write them thus —
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164 MULTIPLES OF NUMBERS.
504 = 23 X 32 X 7
756 = 22 X 33 X 7
924 = 22 X 3 X 7 X H
1176 = 23 X 3 X 72
(r.| Then, since 1176 contains all the factors of each of the othe^
nuaibers, except 11 X 32, the L. C. M. must equal 1176 X U X 32; or
if divided by 1176, the quotient would be 11 X 32, or 99.
{d.) Again. Since 924 contains all the factors of each of the othei
numbers, except 2 X 7 X 32, the L. C. M. must equal 924 X 2 X 7
X 32; or, divided by 924, the quotient must be 2 X 7 X 32, or 126.
{e.) Similar considerations will enable us to obtain the L. C M. from
756 and from 504. A little practice will enable the pupil to determine
at a glance from which number the L. C. M. can best be obtained.
1. What is the L. C. M. of 24 and 36 ?
^M</on. — Observing that 36 = 3 X 12, and 24 = 2 X 12, it will
at once be seen that the L. C. M. may be found by multiplying 36 by 2,
or 24 by 3, and is 72.
2. What is the L. C. M. of 21, 35, and 56 ?
Solution. — It is obvious that .56 contains all the factors of 21 and 35,
except 3 and 5.* Hence the L. C. M. = 56 X 5X3 = 840.
Second Solution, — Since 21 contains all the factors of 35 and 66,
except 5 and 8, the L. C. M. must be 21 X 5 X 8 = 21 X 40 = 840.
Note. — In the above example, it is better to begin with 21 than 56,
because it is easier to multiply 21 by 8 X 5, or 40, than to multiply 56
by 3 X 5, or 15.
3. What is the L. C. M. of 8 and 9 ?
Solution. — Since 8 and 9 have no common factors, their L. C. '^L
must be their product, which is 72.
(/.) What is the least common multiple —
4. Of 12 and 25?
5. Of 32 and 40?
6. Of 16 and 27?
7. Of 12 and 30?
8. Of 24 and 35 ?
9. Of 42 and 70?
15. What is the L. C. M. of 15, 18, 30, 12, and 36 ?
♦ For 66 = 8 X 7, 35 = 5 X 7, and 21 = 3 X 7 ; and 8, 5, oni 8 aft
prime to each other.
10. Of 4, 6, and 12?
11. Of 15, 25, and 35?
12. Of 6, 8, and 9?
13. Of 4, 9, and 25 ?
14. Of 14, 21, ajid 24 ?
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MULTIPLES OF N13IBERS. 155
Solution. — Since 36 is a multiple of 18 and 12, and 30 is a multiple
of 15, the L. C. M. required must be the L. C. M. of 80 and 36, which
equals 5 X 36, or 6 X 30 = 180.
(ff.) What is the least common multiple —
16. Of 3, 6, 12, and 18?
17. Of 4, 9, 12, and 18?
18. Of 5, 3, 6, and 15 ?
19. Of 12, 24, 36, and 72 ?
20. Of 14, 15, 18, 30, 36, and 42 ?
21. Of 9,10, 11, 12, and 18?
22. Of 98, 154, 198, and 284?
23. Of 72, 108, 180, and 252 ?
24. Of 306, 408, 612, and 1020 ?
25. Of 1548, 2322, 2580, and 3870 ?
26. Of 1872, 4212, 6318, and 8424?
118. When Factors cannot easily he found.
(a.) Since two nambers can have no other common factois
than those of their greatest common divisor^ it follows that the
least common multiple of any two numbers may be found by
multiplying one of them by the quotient obtained by dividing
the other by their greatest common divisor. When prime to
each other, their L. 0. M. will be their product.
(5.) This principle may be advantageously applied when we
wish to find the least common multiple of numbers the factors
of which cannot easily be found.
1. What is the L. 0. M. of 7379 and 9263 ?
Solution, — As these nxmibers cannot readily be divided into their
factors, we first find their G. C. D., which is 167. Diyiding 7879 by 167
IpTes 47 for a quotient, which most contain all the factors of the L.C.M.
that are not found in 9268. Hence the L.O.M. must equal 9268 x 47
IB 486861.
(c.) What is the least common multiple —
2. Of 5207 and 7493?
8. Of 2993 and 8651?
4. Of 8837 and 7471?
5. Of 3901 and 9047 7
6. Of 6659 and 8083?
7. Of 7379 and 9263?
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156 FRACTIONS.
8. What is the L. C. M. of 3763, 5183, and 7261 ?
Suggettion. — First find the L. C M. of any two of the namben, as
3763 and 5183, and then the L. C. M. of this result and the remaining
Dambcr.
fd,) What is the least common multiple —
9. Of 2881, 4171, and 9313 ?
10. Of 2419, 4661, and 5609 ?
11. Of 3811, 6031, 7519, and 7661 ?
12. Of 8381, 9809, 7361, and 6817 ?
SECTION X.
FRACTIONS,
11 9. Introductory.
(a.) If an apple, an orange, a line, or any other thing, or any nmii-
ber, be divided into two equal parts, the parts are called halves of the
appUy orange, line, or of whatever may have been thus divided. If two or
any number of things of the same kind are each of them divided into
two equal parts, the parts will, as before, be called halves of a thing of
that kind, and there will be as many times two halves as there are
things divided.
(6.) If any thing should be divided into three equal parts, the parts
would be called thibds of the thing. If the thing divided should be an
apple, the parts would be thirds of an apple. If each of several apples
should be divided into three equal parts, all the parts, or any portion of
them, would still be called thirds of an apple. Again; if I should cut oat
of an apple such a part as would be obtained by dividing it into three
equal parts, and then cut from another apple such another part, these
parts would still be called thirds of an appk, although one of them came
from one apple and one from another. And so, however many or few
such parts we may take from the same apple, or different apples, they
would all be called thirds of an apple,
(c.) Hence, to have a third or thirds of any thing, quantity, or number,
it is only necessary to have one or more such parts as would be obtained b^
dividing the thing, quantity, or number into three equal parts.
(d.) Sach considerations, extended, give the definitions of the fol
lowing article.
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FBACriONS. 157
130* Definitions of Halves, Thirds j S^c.
(a.) Such parts as are obtained by dividing any thing or
number into two equal parts, are called halves of that thing
or number. One such part is called one half of it ; two such
parts are called two halves of it ; three such, Uiree halves of
it; &C.
(h.) Such parts as are obtained by dividing any thing or
number into three equal parts, are called thirds of that
thing or number. One such part is called one third of it ; two
such, two thirds ; three such, three thirds ; four such, four
thirds; Sec
(c.) In like manner, such parts as are obtained by dividing
any thing or number intoybur equal parts, are called fourths
of that thing or number ; such as are obtained by dividing it
into five eqiud parts, are called fifths ; into six, are called
sixths; into seven, sevenths; into eight, eighths; into
nine, ninths ; into ten, tenths ; &c
(<L) 1. What are sevenths ?
Anstuer. — Sevenths of any thing or number are such parts as are
obtained by dividing it into seven eqaal parts.
2. What are fifths ?
3. What are thirds ?
4. What are thirteenths ?
5. What are fourths ?
6. What are ninths ?
7. What are tenths ?
8. What are halves ?
9. What are twenty-firsts?
(e.) From the method of obtaining halves, thirds, &c., it
follows that —
Halves are equal parts of such hind that two of them would
eqtud a unit; thirds are equal parts of such kind that three
of them would equal a unit ; S^c,
(/.) Let the pupil now answer all the preceding questions
in this article according to the following model.
What are sevenths ?
Answer. — Sevenths of any thing or number are equal parts of such
kind that it will take seven of them to equal that thing or number.
Note. — The explanations of {d.) and (/) arc alike necessary to a
thorongh understanding of fractions, and the student should not rest sat-
mfied till he has become so familiar with both, as to be able to use either.
14
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'158 FRACTIONS.
131. Fractional Parts,
(o.) Such parts as the above are called fraciional
PARTS, and the arithmetical expressions for them are called
FRACTIONS ; hence, —
1. Fractional parts of any thing^ quantity ^ or number are
such parts as are obtained by dividing it into equal parts.
2. Fractional parts of any thing, quaiitity, or number are
equal parts of such kind that a given number of them toili
eqiuil a unit.
Note. — That from which the fractional parts are obtained is always,
when considering the parts, regarded as a unit, and is called the unit of
ihe fraction. It may be, —
1. A single object, as an apple^ an orange.
2. A unit of measure, as a foot^ a yardy a bushel*
3. The abstract unit, one.
4. A number of single objects or units considered as a collection or
whole, as 6 apples, IS feet, 24 bushels.
5. An abstract number, as 5, 8, 12.
6. A fraction, as |., 2.
When no particular unit is expressed, the abstract unit is the one
referred to.
(b.) It is obvious that the value of any fractional part de
pends both on the nature of the unit and the number of such
parts which it takes to equal it.
Illustrations. — 1. If a large apple and a small one be each divided
into 2, 3, 4, or any other number of equal parts, the parts of the first
will be larger than the corresponding parts of the second.
2. If several apples of the same size are divided, one into two equal
parts, another into three, another into four, another into five, &c., tha
parts of the first apple will be larger than those of any other, the parts
of the second will be smaller than those of the first, but larger than
those of any other, &c.
(c.) K two equal units are divided, one into two equal
parts, and the other into four, each part of the first will be
equal to two parts of the second; if one be divided into two
equal parts, and the other into three times as many, each part
of the first will be equal to three parts of ihe second ; andt
generally, each part obtained by dividipg a unit into any nur.
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FRACTIONS. 159
bcr of equal parts, will be ttcice as large as each would be if
the unit had been divided into twice as many equal parts ;
three times as large as if divided into three times as many ,
four times as large as if divided into four times as many; SfC,
(d,) In other words, while the unit of the fraction remains
the same, each half will be twice as large as each fourth, 3
times as large as each sixth, 4 times as large as each eighth,
d:c. ; each third will be twice as large as each sixth, 3 times
as large as each ninth, 4 times as large as each twelfth, &c. ;
each fourth will be twice as large as each eighth, 3 times as
large as each twelfth, &c.
(e.) From these considerations, it follows that in order
that fractional parts may be of the same denomination, it ia
necessary, —
1. That they should be parts of the same unit, or of equal
units ;
2. That they should be obtained by dividing each unit into
the same number of equal parts.
1S3. Definitions. — Method of writing Fractions, S^c.
(a,) A FRACTION expresses the value of such parts as are
obtained by dividing a unit into equal parts ;
Or, by definition second, —
A FRACTION expresses the value of such equal parts, that a
certain number of them will equal a unit,
(b.) In writing fractions by figures, two numbers are neces-
sary — one to indicate the number of parts into which the
unit is divided, or (which is the same thing) the number of
Buch parts which it will take to equal the unit ; the other to
Indicate how many of the parts are considered.
(c.) The first of these is called the denominator, because
I* indicates the denomination of the parts. The second is
called the numerator, because it numbers the parts.
(d,) The numerator is usually written above the denomi-
nator, and separated from it by a line.
Illustrations. — Five sixtlts is written ^, 5 being the nnmerator, and 6
the denominator.
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160 FRACTIONS.
Seventh eightlis is written |, 7 being the numerator, and 8 the denoin
inator.
34 twenty-firsts is written .^, 34 being the numerator, and 21 tha
denominator.
{e.) Write the following fractions by figures, and tell the
numerator and denominator of each.
1.
3 fourths.
6. 1 third.
2.
9 twenty-seconds.
7. 18 thirteenths.
3.
7 halves.
8. 5 elevenths.
4.
9 seventeenths.
9. 15 thirty-firsts.
5.
22 ninths.
(/.) In DECIMAL FRACTIONS the numerator only is writ-
ten, the denominator being determined by the position of the
decimal point. (See 22^ a.)
133* Exercises in explaining Fractions.
1. Explain the fraction f.
Answer. — Five nintlis, or five ninths of one, expresses the value of
five such parts as would be obtained by dividing a unit into nine equal
parts.
Another Form of Ansiver. — Five ninths, or five ninths of one, ex-
presses the value of five equal parts, such that nine of them would equal
a unit.
2. Explain the fraction .07.
First Form of Answer. — Seven hundredths, or seven hundredths of
one, expresses the value of seven such parts as would be obtained by
dividing a unit into 100 equal parts.
Second Form of Answer. — Seven hundredths, or seven hundredths of
one, expresses the value of seven equal parts of such kind that 100 of
them would equal a unit.
Explain the following fractions according to the first fom*
of answer, and afterwards according to the second.
3.
f
6.
.09.
9.
*.
4.
H'*
7.
.6.
10.
.008.
5.
??•
8.
.27.
11.
^'
2B fortf/'firats, not 25 fort}-K>nethB, nor 25 forty-onei»
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FRACTIONS.
I
12.
if
16. .427.
20.
«•
IS.
■^•
17. .0628.
21.
n-
14
i§.
18. .000276.
22.
.0107.
15.
m-
19- H-
23.
.002006.
161
1241. Various Kinds of Fractions,
((u) A simple fraction is one which has but one numerator
tnd one denominator, each of which is a whole number, ai |,
(b,) Simple fractions may be proper or improper.
(c.) A proper fraction is a fraction whose numerator is
less than its denominator, as t\, ^, -ff .
(d,) An improper fraction is one whose numerator is equal
to, or greater than, its denominator, as |, -^, -if^.
(e.). A mixed number is a whole number and a fiaction, as
2^, which is read two and one fiftk ; 5^, which is read five
and three sevenths,
1. Which of the fractions in 133 are propei ?
2. Which are improper ?
(/.) A proper fraction is less than 1, because it expresses
less parts than it takes to equal a unit.
(^.) An improper fraction is equal to, or greater than, 1,
because it expresses as many or more parts than It takes to
equal a unit.
(A.) An improper fraction is so called because it expresses
a value, a part or all of which may be expressed in whole
numbers.
13«S. Illustrations of Operations on Fractions,
Fractions may be added, subtracted, multiplied, and divided,
•8 whole numbers are.
Thus, f + f- = f-, jnst as 5 days + * days = 9 days,
f + f = "V^, just as 7 qts. + 3 qts. = 10 qts.
« 19 twenty-aecondaf not 13 twenty-twos.
14*
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132 FRACTIONS.
f f + ii = Hi just as 23 lbs. + 21 lbs. = 44 lbs.
Again. :J + J ^^^^ "*o**® equals J, or f , than 3 qts. 4" ^ pt =» 4 qts ,
or 4 J t.
xi + A no n»orc equals if, or ^|, than ll5 + 53=16g,or
16 3.
Again, f — § =i f , just as $8 — $3 = $3, or 8 lbs. — 3 lbs. »«
6 lbs.
}?- — -} ^ = ^, just as 15 rods — 13 rods = 2 rods.
f I J — ^1^ s jjj^, just as 358 apples — 183 apples == 175 apples.
Again. J — j no more equals f , or f, than 3 pints — 1 gill sss 2
pints, or 2 gills.
^ — I no more equals y, or ij, than 5 apples — 4 pears == 1 apple, or
1 pear.
Again. 5 times J = ^^ just as 5 times 3 apples = 15 apples.
8 times i^ = f |. just as 8 times 11 oz. = 88 oz.
9 times if fj = ^^f^N^t jwst as 9 times 1387 cubie inches = 12483
cubic inches.
Again, f are contained 3 times in f , just as 2 days are contained 3
times in 6 days.
■^ are contained 7 times in f j-, just as 3 grains are contained 7 times
in 21 grains.
Th *r® contained 18^ times in iff, just as 7 apples are contained
18^ times in 129 apples.
Again, i of | = f , just as i of 8 apples = 2 apples.
i of §i = -jy , just as i of 24 bushels = 4f bushels.
i of iilf = jriT ) j°st as i of 1484 miles = 164f miles.
130* deduction of Whole or Mixed Numhers to Improper
Fractions,
The value of any whole or mixed numher may he ex-
pressed hy an improper fraction.
1. Reduce 8 to fifths.
Solution, — Since I = f , 8 must equal 8 times f, or ^. Therefore,
8 = 4^.
2. Reduoa 9f to sevenths.
Solution. — Since 1 = f, 9 must equal 9 times 7, of -\^, and y added
•TO ^. Therefore, 9f = V"-
3. Reduce 4 to thirds | 4. Reduce 5| to thirds.
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FRACTIONS.
163
8. Reduce 9f to fourths.
9. Reduce 430f to fourtha.
5. Reduce 6f to eighths.
6. Reduce 8 to eighths.
7. Reduce 9f to fifths.
First Solution. — Since 1=4 fourths, 439 most eqaal 439 times 4
fourths, which is equivalent to 4 times 439 fourths, or 1756 fourths, and
3 fourths added are 1759 fourths. Therefore, 439f = -^^.
Second Solution. — Since there are 4 fourths for each unit, there most
be 4 times as many fourths in any number as there are units. Hence,
439 s= 4 times 439 fourths, or 1756 fourths, and 3 fourths added arc
1759 fourths. Hence, 439| = i^.
Note. — Compare these reasoning processes with those given on tho
85th page.
10. Reduce 578f to ninths.
11. Reduce 14762^ to sevenths.
12. Reduce 74968^^^ to seventeenths.
13. Reduce 6483f f to forty-sevenths.
14. Reduce 4386^f to twenty-eighths.
15. Reduce 427jf f to two hundred and seventy-thirds.
16. 9 = how many times J ? *
First Solution. — Since 1 = J, 9 must equal 9 times J, or •^, and 1
fourth is contained in 36 fourths 36 times. Hence, 9 = 36 times ^.
Second Solution. — Since 1=4 times 5-, or |-, 9 must equal 9 times
4 times i, or 36 times i. Hence, 9 = 36 times i.
17. 7 = how many times ^ ?
18. 11 = how many times -J ?
19. 49 iz: how many times ^?
20. 487 = how many times ^ ?
21. 9 = how many times -j\y ?
22. 7 = how many times .1 ^
23. 13 = how many times .01 ?
24. 648 = how many times .0001 ?
1 27. Heduction of Improper Fractions to Whole or Mtxeti
Nuinhers,
The value of any improper fraction may be expressed by a
whole or mixed number.
* It will be seen by the solutions that this question is just equivalent to
•« 9 =« how many fourths ? "
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164 TRACTI0X8.
1. Reduce V ^ ^^^^
Since f = 1, ^ must equal as many ones as there are times 8 in
43, which are 5 3^ times. Hence, ^ s= 5f.
NoTB. — The preceding question is precisely like this : ** 48 qts. ssa
hoT many pecks ? ** as the following solution will show. " Since 8 qts.
^ I pk., 43 qts. will equal as many pecks as there are times 8 in 43,
which are sf times. Hence, 43 qts. = 5$ pks.*'
2. Reduce ^ to ones.
8. Reduce ^ to ones.
4.- Reduce J^f^ to ones.
5. Reduce ^^ to ones.
6. Reduce ^||* to ones.
7. Reduce f ^f f to ones.
8. Reduce ^ to ones.
9. Reduce f J to ones.
10. Reduce -If ^ to ones.
11. Reduce ^^ to ones.
12. Reduce ^^^ to ones.
13. Reduce ^ff * to ones.
138* Miscellaneous Questions involving Fractions.
1. How much will 227 yards of cloth cost at J of a dollar
per yard ?
Solution. — Since 1 yard costs f of a dollar, 227 yards will cost 227
times J of a dollar, equivalent to 3 times ^f^ of a dollar, which, found
by multiplying 227 by 3, is ^f ^ of a dollar, equal (by 127) to 170^
dollars.
Note. — Compare the preceding example and solution with the fol-
lowing : —
How many bushels of grain in 227 bags, each holding 3 pecks ?
Solution. — Since 1 bag holds 3 pecks, 227 bags will hold 227 times
8 pecks, equivalent to 3 times 227 pecks, which, found by multiplying
227 by 3, is 681 pecks, equal to 170 bushels, 1 peck.
2. What will 493 lbs. of tea cost at J of a dollar per lb. ?
3. What will 1286 bu. of apples cost at f^ of a dollar
per bu. ?
4. What will 347 gal. of burning fluid cost at |J of a dol-
lar per gal. ?
5. How many acres in 169 house lots, each contmning f|
•f an acre ?
6. How many acres in 3 lots, each contuning |^§f|^ acres?
7. How many quarts in 9 boxes, each holding |4 of •
quart ?
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FRACTIONS. 163
8 How many pounds of silver in 28 bars, e&cli weigliing
I^H of a pound ?
9. How many tons of hay in 37 loads, each containing
iUi of a ton ?
10. How much will 7 acres of land cost at $275 J per
acre?
Note. — Compare the last example with this : " How much corn is
7 bins, each holding 275 bushels, 3 pecks ? "
11. How much will 6 acres of land cost &i $493 J per
acre?
12. If a ship sails uniformly at the rate of 179-f^ miles
per day, how far will she sail in 5 days ?
13. If a steamship sails uniformly at the rate of 4179f||>
rods per hour, how many rods will she sail in 85 hours ?
14. How many baskets, each holding J of a pe^k, can be
filled from 82^ pecks of corn ?
Solution. — Since J of a peck will fill one basket, 82 § pecks will fill
as many baskets as there are times J in 82 f. But 82f = -^1^, which
contains i as many times as 659 contains 7, which is 94^ times. Hence,
94-^- baskets can bo filled.
Note. — Compare the above question and solution with the follow«
ing : " How many baskets, each holding 7 quarts, can bo filled f-^m
82 pk. 3 qt. of com ? "
Solution, — Since 7 quarts will fill one basket, 82 pk. 3 qt. will fill as
many baskets as there are times 7 qt. in 82 pk. 3 qt. 82 pk. 3 qt. = 659
qaarts, which contains 7 quarts as many times as 659 contains 7, which
is 947 times. Hence, 94^- baskets can be filled.
15. How many vest patterns, each containing f of a yard,
can be cut from a piece of vesting containing 25^ yards ?
16. How many books, at f of a dollar a piece, can be
bought for 232^ dollars ?
17. How many baskets, each holding ^ of a bushel, can
be filled from 453/2- bushels of apples ?
18. How many barrels, holding 2^ bushels each, can be
filled from 59f bushels of apples ?
Note. — Compare the last question with, — " How many jugs, hold*
log 2 gal. 1 qt. each, can be filled from 59 gal. 3 qt. of fluid 1 "
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166 FRACTIONS.
19. How many spoons, each weighing 7^ dwt, can b6
made from 328J dwt. of silver ?
20. How many yards of cloth, at 2^§ dollars per yard, caL
be bought for 89f J dollars ?
21. How many days, at $2^^ per day, must a man labor
to earn $237^1 ?
22. A man, who owned 247-j^ dwt« of silver, had it made
into spoons, each weighing 4^^ dwt How many spoons did
it make ?
23. What is the sum o£ ^^ + ^ + -^ + ^ + ^ +
Note. — Compare the above example with 11 in. + 7 in. + 9 in. -f-
4 in. + 7 in. -|- 5 in. = 43 in. = 3 ft. 7 in.
What is the sum —
24. Oft + * + i + t + *^
25. Of^ + ii + /ir + A + A + «+ff?
26. Of|f + it + ^§ + iJ + fi + ^^ + H?
27. Of 13^ + 13f + 2f + 22f + 43f ?
NoTB. — Compare the 27th example with the following: — 13 ir
8 da. + 13 w. 5 da. -f 2 w. 4 da. -|- 22 w. 6 da. -f 43 w. 2 da. == 95 w.
6 da. To make the resemblance to compound addition still more ap-
parent, we give two forms of writing the work of the 27th example, and
place opposite them the work of the example in this note.
Ones.
Soventha.
w.
da.
13^ == 13
3
13
3
13^ = 13
5
13
5
2f = 2
4
2
4
22f = 22
6
22
6
43f = 43
2
43
2
95f = 95 6 = -4n«. 95 « = .Am.
What is the value —
28. Of 17^ + 18A + 23A + 42^7^ + 271-? ?
29. ' Of 24J + 96J + 14f + 26^ + 55 J ?
* These, like the compound numbers of 56, can best be reduced at
they are added ; thus, j^ + y^g- (by adding one of the -^ to fj^) = 1^ 5
•ad 1^ -f- ^ (by adding 3 of the ^ to the i^) = 2^^ ; &c.
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ruACTiONS. 167
80. Of 237iJ + 496^ + 849^^ + 591jjg + 438ig ?
31. Of I — I ?
32. Ofil^-T^^?
83. Of^VifeV^-2*ff%VA?
34. Of 24/^ — 19|i?
Note. — Compare the 34th with the following : — What is the faint
cf 24 qr. 7 lb. — 1 9 qr. 21 lb.
Bj writing the work of the 34th example in the following form, and
placing opposite it the work of the example in this note, the resem*
blance of the two is very clearly exhibited.
Ones. Twenty-fiftbs. Qx. lb.
24yV = 24 7 . 24 7
. 19||=19 21 19 21
m= 4 n^Aru. 4 U^snAns.
What is the value —
35. Of 247ff — 159if?
36. Of 4327ff — 2589|i?
37. Of 6732f §J J — 4694f ^ ?
38. Of 479yf ly — 214i|f^ ?
39. 0£2SmiH — 2S3um?
40. Of 9243f|Hf — 3842f|fH?
139. One Number a Fractional Part of another.
What part of 4 is 1 ?
Answer. 1 is one fonrth of 4, because 4 times 1 are 4."*
What part —
1. OfGisl? I 3. Of 9 is 1? I 5. Of 16 is 1?
2. Of 2 is 1 ? 4. Of 12 is 1 ? I 6. Of 243 is 1 ?
7. What part of 8 is 3 ?
Solution. — Since 1 = ^ of 8, 3 mast eqnal f of 8.
What part —
8. Of 9 is 4 ?
9. Of 11 is 7 ?
10. Of 3 is 2 ? I 12. Of 6 is 5 ?
11. Of 8 is 5? 13. Of 13 is4i
* i. e., Because 1 taken 4 times will e<|aal 4
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VRACTIONS.
14. What part of 7 is 9 ?
Solution."' Since 1 « ^ of 7, 9 most eqaal f of 7, or if times 7
What part —
15. Of 9 is 13?
16. Of 10 is 87 ?
17. Of 19 is 89?
18.
19.
20.
Of 47 is 63?
Of 89 is 19 ?
Of 413 is 527?
21. What part of 8 yards is 1 yard ?
Firat Form of Answer, 1 yard = i of 8 yards, because 8 times 1
yard s= 8 yards.
Second Form of Answer, 1 yard is the same part of 8 yards that 1 is
of 8, which is j-. Hence, 1 yard =: ^ of 8 yards.
What part —
22. Of 9 fl. is 1ft.?
23. Of 4m. is 3 m.?
24. Of 41b. is 31b.?
28. What part of f is f ?
Answer, f is the same part of f that 3 is of 4, which is J.
fisfoff
What part —
25. Of 4 cwt is 3 cwt ?
26. Of 10da.is7da.?
27. Of 8yr. i8 5yr.?
Hence,
29. Of lis f?
80. Of TfyiSyfy?
81. Ofilisff?
82. Of T^is^VV?
33. Of T^yiST^?
34. Of V^is^?
35. Of fist?
86. Of ft is «?
87. What part of 1 bushel is 3 pecks ?
1 bu. s= 4 pk., and 8 pk. = J of 4 pk. Hence, 3 pk. ■■
Solution.
I of 1 ba.
What part —
88. Of 1 m. is 5 fur. ?
89. Of lib. is 7 5?
40. Of lA.is3R.?
41. Of 1yd. is 2ft.?
42. Of 1 fti. is 1 in. ?
43. Of IT. is 13 cwt?
44. What part of 8 da. is 1 w. ?
Solution, 1 w. = 7 da., and 7 da. = f of 3 da., or 2 J times 3 da.
What part —
45. Of 5 d. is 1 s. ? I 47. Of 13 dr. is 1 oz. ?
46 Of 7 s. is £1? I 48. Of 9 dwt. is 1 ox. ?
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1G9
49. Of 11 S is 1 lb. ? I 50. Of 8 da. is 1 w. ?
51. What part of f is 1 ?
Solution, 1 = f , and f is the same part of i that 6 is of 5, which li
f . Hence, 1 «= f of f , or 1^ times ^.
What part —
52. Ofvft:isl? 57. Of V is 1 ?
53. Of fisl? 58. Of^isl?
54. Offisl? 59. Of if is 1?
55. OfV^isl? 60. 0{^t\isl?
56. Of^isl? 61. OfyfJ^isl?
62. What part of If is 2f ?
Solution, if = V;2f = V;andV = ftof-V- = lH timM
63. What part of 1 m. 3 fur. is 2 m. 5 fur. ?
Solution. 1 m. 3 fur. = 11 fur. ; and 2 m. 5 fur. &= 21 for. ; 21 ftir. ■■
H of 11 ^M or li^ times 11 fur.
What part —
64. Of Id. 3qr. islld.lqr.?
65. Of If is 11 J?
66. Of 8w. 3da. is2w. 4da.?
67. Of 8^ is 2f?
68. Of 5 da. 7 h. is 3 da. 11 h. ?
69. Of 5/yis3ii?
70. Of 11 h. 27 m. is 21 h. 88 m. ?
71. Of 15pk. is3pk. 7qt?
72. What part of 1 lb. is 3 oz. 5 dwt. 17f gr. ?
Suggestion. — Kednce 1 lb. and also 3 oz. 5 dwt. 17| gr. to the lowMl
ienomination mentioned, i. e., to fifths of a grain.
What part —
73. Of 1 gal is 3 qt. 1 pt. 3 gi. ?
74. Of £1 isl7s. lid. 2qr.?
75. Of 1 w. is 5 da. 11 h. 35 m. 29f sec ?
76. Of 1 T. is 13 cwt 2 qr. 19 lb. 13 oz. 11 J dr. ?
77. Of 4 yd. 2 ft. 7 in. is 1 yd. 1 ft. 11 in. ?
78. Of 17A. 8R. 16sq. rd. is2A. IR. 28 8q.rd.P
15
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FRACTIONS.
130* Other Methods of expressing the Value of a Fraction.
(a.) 1. Explain the fraction f .
Ansiver, f = 3 times |-, as f of a nambcr = 3 times i of th«
number.
In the same way explain the meaning —
2. Off. 5. OfiJ. 1 8. Of .4
8. Off. 6. Of t\,V 9. Of .007
4. Of |. 7. Of .15 10. Of .0346
(b.) Explain each of the above by the following method.
J, or f of 1, s=s i of 8, as f of ft number = J ^f 3 times the num
ber. For if each of 3 equal things should be divided into 4 eqaal parts,
and 1 of the parts of each taken, the result would equal f of 1.
NoTB. — This may be illustrated to the eye, by taking 3 equal lines,
^^^^^^^^^^ and dividing them into 4 equal parts, arrang^ thus
='=== one of the parts will then contain i of 3 lines, which,
M will be seen, is equal to j of a line.
11. To what simple fraction is jt of d equal ?
Answer, i of 3 = f of 1 = f .
(e.) To what simple fraction is each of the following
equal?
12.
*of 2.
16.
i of 11.
20.
tV of 832,
18.
+ of 4.
17.
A of 4.
21.
1 of 98.
14.
i\ of 6.
18.
■iV of 15.
22.
aV of 493.
15.
iof 7.
19.
i of 49.
23.
jV of 861.
131* To find a Fractional Part of a Number.
1. What is i of 5459 ?
First Solution, i of 5459 = 7 times i of 5459 ; i of 6459, ]
by dividing by 8, is 682 J, and 7 times 682| = 4776t.
Second Solution, i of 5459 = j of 7 times 5459 ; 7 times 5459 ■■
t8S13, and i of 38213 == 4776f .
WSITTBN WORK.
By First Solution.
8 ) 5459
682|
7
JSy Second Solution*
5459
7
4776f s Ans,
8 > S821S
4776f » ^fif.
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FRACTIONS. ]71
The following forms of writing the work exhibit very deariy tht
rc(*)on for every step which is taken. The letters a, b, &c^ arc used m
explained in the note on the 88th page.
First Form, Second Form,
9 ■= 5459 a =s 5459
i of a = b = 682f 7 X a = b = 38213
7Xb = }ofa = 4776| i of b == J of a = 4776|
Note. — It will be seen that by both methods of eolation, we multi-
ply by the numerator, 7, and divide by the denominator, 8. The only
difference between them is, that by the first we divide before we muUiply^
while by the second we multiply before we divide. The first method
ms olives somewhat smaller numbers than the second.
What is the value of each of the following ?
2. f of 3843. I 4. f of 7148. I 6. fj of 46935.
3. {of 7987. I 5. .{-i of 58437. 1 7. ^fj of 875 16.
8. What is the value of .0014 of 16.427 ?
First Solution. .0014 of 16.427 == 14 times .0001 of 16.427 ; .0001
of 16.427, found by removing the point four places towards the left, it
0016427, and 14 times this is .0229978 = Ans.
The work may be written in full after either of the following models
First Model,
a = 16.427
.0001 of a = b == .0016427
14 timc» b = .0014 of a = .0229978 = Ans.
Second Model
10000 ) 16.427
.0016427
14
.0229978 a Ans.
Second , Solution. .0014 of 16.427 = .0001 of 14 times 16.427; 14
times 16.427 = 229.978, and .0001 of 229.978, found by lemoving tha
point four places towards the left, is .0229978 == Ans.
The work may be written in full after either of the following models
First Model.
a= 16.427
14 times a » b = 229.978
.0001 of b = .0014 of a = .0229978
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Second Modd,
16.427
14
10000 ) 229.978
.0229978
NoTB. — It will be seen that bj both solutions, we mvlUply by M
numerator^ 14, and divide by the denomincUor, 10000; i. e., we mnltiplj
by 14, and remove the point four places towards the left. The only dif-
ference between them is, that in the first we remove the point before ive
multiply f while in the second we remove it afterwards. Some of the above
written work might have been avoided by multiplying by 14, and chang-
ing the position of the point at the time of writing the product, thus : — •
16.427
.0014
.0229978
In like manner we may get .07 of a number by multiplying by 7,
and removing the point two places towards the left ; we may get 5.2 * of »
number by multiplying by 52, and removing the point one place towards
the left ; &c
What is the value of each of the following ?
9.
.03 of 2589
14.
.006 of .006
10.
.005 of 3.7479
15.
2.5 of 2.5
11.
.28 of 437.96
16.
.372 of 5.46
12.
4.2 of 67.93 1
17.
.0004 of 200
13.
.25 of .25
18.
.006 of .006
133. To multiply hy a Fraction.
1. What is the product of 865 multiplied by ^ ?
Suggestion. — The product of 865 multiplied by ^ equals •? timet
865, which is the same as ^ of 865, and may be found by methods be-
fore explained.
Note. — This follows from the definition of multiplication — A
process by which toe ascertain how much any given number will amount to^
if taken as many times as there are units in some other given number. In
multiplying by more than a unit, then, more than once the multiplicand
.8 taken, while in multiplying by less than a unit, less than once the mul-
* It will be more convenient, for purposes of explanation, to regard this
as an improper fraction than as a mixed number.
f The number of which a fractional part is taken may be regarded eithef
as a mixed number or an improper fraction.
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FRACTIONS. 178
tiplicand (i. e., some fractional part of it) is taken ; the prodact being
always the same part of the multiplicand that the multiplier is of unitj.
When, therefore, the multiplier is greater tlian unity, the prodact will
be greater than the multiplicand ; and when the multiplier is less than
unity, the product will be less than the multiplicand. In multiplying
by ^ j- of once the multiplicand is taken ; in multiplying by f, i of
once the multiplicand is taken ; &c.
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174 FRACTIONS.
Second Form.
a= 29i8
— 1
3 X a = b = 8844
iofa==C'-=: 327^
4 X c = d = 13lo| = J of a
#b4-c + d = 8^ + a= 10481 J = Ans,
KoTE. — The second of the abore forms is preferable to the firnii
inasmach as it presents to the eye, in a compact and convenient form,
all the numbers which it is necessary to use, and avoids all side work.
The references by means of letters may be omitted as soon as the pro*
cesses and explanations are familiar.
What is the product -
9. Of 8^ times 6794 ?
10. Of 26§ times 2579 ?
11. Of 43J times 45687 ?
12. Of 8f times 85476?
13. Of 25T*r times 2764?
14. Of 698i times 29679 ?
133. Practical Problems.
1. How much will { of ah acre of land cost at $478.36
per acre ?
Solution. — If 1 acre costs $478.36, J- of an acre will cost i of
$478.36, which, found in the usual way, is $41 8.56 j-.
First Form. Second Form.
8 ) $478.36 a = $478.36 a= cost of 1 acre.
• ^ of a = b = $ 59.79^ = cost of i of an A.
*59.79i- 7 X b == $418.56| = cost of | of an A.
7
$418.56^ = Ans
Note. — The second form is better than the first, when a full state-
ment and explanation of the successive steps are required ; bat in other
cases the first is preferable.
2. How much will .9 of an acre ^f land cost at $394.26
per acre ?
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FRACTIO.^S. 175
Solution. — If 1 acre costs $394.26, .9 j>f an acre will cost .9 of
f394.26, which, foand as before explained, is $354,834. The following
exhibits the written work.
Full Form. Abbreviated Form,
a = $394.26 = cost of I acre. $394.26
.1 of a = b = $ 39.426 = cost of .1 of an A. .9
9 X b = $354,834 = cost of .9 of an A.
$354,834 = Ans.
3. How much is ^ of a vessel worth, if the whole vessel is
worth $29951.88 ?
4. I bought f of an acre of land at $397.36 per acre.
How much ought I to pay for it ?
5. If a bushel of tomatoes weighs 48.75 lb., what will ^^
of a bushel weigh ?
6. What will .4 of a ton of iron cost at $47.93 per ton ?
7. What will .8 of a bag of coffee cost at $27.94 per bag?
8. Mr. Foster bought 4377 bushels of com, and sold .37
of it to Mr. Gardiner, .43 of it to Mr. Calder, and the rest to
Mr. Godfrey. How many bushels did he sell to each ?
9. What will ^ acres of land cost at $126 per acre ?
10. What will 32ff acres of land cost at $139.50 per
acre?
11. The ship Surprise is valued at $25427. Joseph Ward
owns f of her, and Samuel Lowe owns the remainder. What
is the value of Mr. Ward's share ? Of Mr. Lowe's ?
12. Tlie brig Maria sailed from Philadelphia to Boston,
with 207 T. 13 cwt. 2 qr. 19 lb. of coal, f of which was landed
at one wharf, and the remaindei at another. What was the
weight of that landed at each wharf?
13. If John walks ^ as fast as George, how far will John
walk while George is walking 97 m. 6 fur. 37 rd. 2 yd. ?
14. A company of 9 California gold diggers found in 1
month 37 lb. 4 oz. 16dwt. 17 gr. of gold, which they divided
equally. What was the share of each ?
15. An English ship, valued at £8743, carries a «irgo
which is worth } as much as the ship. How many £, s., and
d. is the cargo worth ? In a storm the sailors were obliged
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176 FRACTIONS,
to throw ^ of llie cargo overboard. What was the value #»f
what they threw overboard ?
16. A merchant bought 17 T. 14 cwt. 2 qr. 23 lb. 10 oz. of
sugar. He sold f of it to one man, ^ of the remainder to
another, and what there was left to another. What was the
weight of that which he sold to the first man? To the
second ? To the third ?
Suggestion. — After selling ^ of the sngar, he must have had -f of it
left, and -^ of this, or what he sold to the second man, must bo -J- of f)
or f of the original quantity ; and the remaining ^ of -f , or -f of the
original quantity, mast have been sold to the thinl man.
Or, the share of the third man may be determined thus : Since the
first man took ^ of the sugar, and the second ^, both took -f of it, which
would leave j- for the third man.
17. A father, dying, left to his son an estate valued at
$87646.44. The hrst year afler the son came into possession
of the property, he spent ^ of it in dissipation, the second year
he spent f of the remainder, and at the end pf the third year
he was penniless. How much did he spend in the first year ?
How much in the second year ? How much in the third ?
Suggestion. — Compare this example with the last
18. If Mr. Smith uses 9.678 tons of coal per year, and
Mr. Parkhurst uses .7 as much, how many tons does Mr.
Parkhurst use ?
Note. — The term per cent is often used in arithmetic, and in busi-
ness transactions, instead of one-hundredths. Thus, 6 per cent has the
same meaning as 6 one-hundredths. 8 per cent == .08 = yS'Tr*
5 per cent of 340 acres = .05 of 340 acres.
6 per cent of $86.38 = .06 of $86.38.
t Every number is 100 per cent ol itself.
19. A teamster carted 7486 bushels of potatoes to a rail-
road depot, receiving in payment 9 per cent of them. How
many bushels did he receive ?
20. A man who owned 378.37 aeres of land gave 8 per
cent of it to liis son. How nimiy acres did he give to hit
son?
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FRACTIONS. 171
21. A city merchant agrees to sell for a country farm*ii
whatever produce the latter may send to him, on condition
that he shall receive for his trouble 5 per cent of the money
he receives for the produce. Under this arrangement he sells
produce to the value of $1768.37. What ought he to receive
for liis trouble ?
Note. — A merchant who makes it his business to buy and sell
goods for others is called a Commission Merchant. The money he
receives for his services is called his Commission. For instance : In
the last example, the merchant receives a commission of 5 per cent on
the valao of the produce he sells. He will deduct this commission from
the amount of the sales before paying the farmer.
22. A commission merchant sold cloth for a manufacturer
to the amount of $7643.79, receiving a commission of 3 per
cent. What did his commission amount to? How much
money ought he to pay the manufacturer ?
23. A commission merchant buys goods for me to the
amount of $387.46, for which he charges a commission of 2
per cent. What will his commission amount to ? How much
money must I send him to pay for the goods and commission ?
24. Mr. Moore borrowed some money of Mr. Boy den,
agreeing to pay him, for each year's use of it, a sum equal to
6 per cent of the money he had borrowed. He borrowed
$125.63, and kept it just one year. How much ought he to
pay for the use of it ? How much, then, ought he to pay Mr.
Boyden in all ?
Note. — Money paid, as in the above example, for the use of money,
is called Interest. The money used is called the Principal. Inter-
est is usually reckoned at a certain per cent of the principal for each
year that it is used. The percentage paid for each year is called the
Rate per Cent. The interest and principal added together form the
Amount.
25. What is the interest of $137.84 for 1 year at 6 per
cent?
26. What is the interest of $487.31 for 6 months at 6 pei
cent per yeai* ?
Suggestion. — Since the rate for 1 year is 6 per cent, the rate for 6
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178 FRACTIONS.
months must be ^ of 6 per cent, which is 3 per cent. The interest for C
months will, therefore, be 3 per cent of the principal.
27. Mr. Adams borrowed of Mr. Wales $718.63, for which
he agreed to pay interest at 6 per cent At the end of 6
months he paid the principal and interest. How much did
he pay ?
28. What is the interest of $47.83 for 4 months at 6 per
cent ?
y
134 • Multiplication and Division of the Numerator.
The foregoing illustrations have shown that multiplying
the numerator of a fraction multiplies the fraction, and that
dividing the numerator divides the fraction. The same thing
may be demonstrated more rigidly, by considering the nature
of fractions, thus : —
Since the numerator of a fraction shows how many parts
are considered, multiplying or dividing the numerator multi-
plies or divides the number of parts considered, without
affecting their size, and hence multiplies or divides the frac-
tion.
1. Explain the effect of multiplying the numerator of the
fraction ^ by 3.
Answer. — Multiplying the numerator of the fraction iV hy 3 gives
Jf for a result, which expresses 3 times as many parts, each of the
same size as before, and is, therefore, 3 times as large. Hence, the
fraction -fj has been multiplied by 3.
See Note after solution of example 8th.
Explam the effect of multiplying the numerator —
2. Of T^ by 2. I 4. Of f by 4.
8. Of /^ by 8. I 5. Of f^ by 11.
6. Of f byO.
7. Of if by 8.
8. Explain the effect of dividing the numerator of the
fraction f by 4.
Solution. — Dividing the numerator of the fraction f by 4 gives f
for a result, which expresses ^ as many parts, each of the same size aa
before, and is, therefore, j- as large. Hence, the fraction f has been
di\ ided by 4.
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FRACTIONS. 179
Note. — The work explained alMjYC is really equivp?'*"!* to 3 times
iV = "H, (just as 3 times 5 apples are 15 apples;) to ^ of f = fi^
(just as i of 8 apples = 2 apples.) The above forms are, howerer,
Decessary, and shoald therefore be mastered.
Explain the efFeet of dividing the numerator —
9. Of f I by 8.
10. Of ^ by 2.
11. Of if by 9.
12. Of if by 3.
13. Of H by 6,
14. Of J J by 12.
l3Sm Multiplication of the Denominator*
\a.) Since the denominator of a fraction shows into how
many parts the unit is divided, or. how many parts equal the
unit, multiplying the denominator must (131) h. and c.)
divide each part, and therefore must divide the fraction.
1. Explain the effect of multiplying the denominator of
the fraction f by 2.
Answer, — Multiplying the denominator of the fraction ihj2 gires
-| for a result, which expresses the same number of parts, each j- as
large as before. Therefore, f = ^ of J, or multiplying the denomina-
tor of f by 2, has divided the fraction by 2.
Explain the effect of multiplying the denominator —
2. Of t by 3. 5. Of ^^ by 7. 8. Of t by 7.
3. Of § by 4. 6. Of J by 2. 9. Of i^ by 10.
4. Of f by 4. I 7. Of f by 3. 10. Of \ by 6.
(5.) Hence, multiplying the denominator of a fraction
divides the fraction^ hy dividing the size of each party with'
out affecting the number of parts considered*
136 • Division of the Denominator*
Since the denominator of a fraction shows into how many
parts the unit is divided, or how many such parts are equal
to the unit, dividing the denominator must (131) h* and c.)
multiply each part, and therefore multiply the fraction.
L Explain the effect of dividing the denominator of the
fraction }i by 4.
Answer. — Dividing the denominator of the fraction -ti by 4 gives
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180 FBACTIONS.
V ^ov a result, which expresses the same number of parts, each 4 timet
as hirge as before. Therefore, -^g^ = 4 times -Ji-, or, dividing the
denominator of -f ^ by 4 has mnhiplied the fraction by 4.
Explain the effect of dividing the denominator —
8. Of ii hy 6.
9. Of ^ by 5
10. Of -^ by 3.
2. Of f by 3. 5. Of 2^ by 5.
8. Of I by 4. 6. Of if by 12.
4. Of T^y by 2. 7. Of ^ by 36.
Hence, dividing the denominator of a fraction tntdtiplies
the fraction^ hy mvltiplying the size of each part without
affecting the number of parts expressed.
137* Recapitulation and Inferences.
(a.) Multiplying the numerator multiplies the fraction, by muUtplying the
nvmber of parts considered without affecting their size.
{b.) Dividing the numerator divides the fraction, hy dividing the ntim&er
rf parts considered wit/iout affecting their size.
(c.) Multiplifing the denominator divides the frojdion, hy dividing each
part without affecting the number of parts considered.
(d.) Dividing the denominator multiplies the fraction, hy multiplying each
part without affecting the number of parts considered.
(e.) Hence, 1. A fraction may be multiplied either by multiplying the
numerator or by dividing the denominator,
2. A fraction may be divided either by dividing the numerator or by
multiplying the denominator.
S. Multiplying both numerator and denominator of a fraction by any
lumber both multiplies and divides the fraction hy that number, and, there'
fore, does not alter its value.
4. Dividing both numerator and denominator of a fraction hy the same
number both divides and multiplies the fraction by that number, and, there-
fore, does not alter its value.
138 • Multiplication and Division of both Numerator and
Denominator by the same Number.
1. Explain the effect of multiplying both numerator and
denominator of the fraction f by 6.
Answer. — Multiplying both numerator and denominator of the i
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FRACTIONS. 181
ikm 4* by 6 gives f ^ for a resalt, which expresses 6 times as manj
parts, each ^ as large as before. Therefore, the value of the fraction is
unaltered, and 4^ == f J.
Explain the effect of multiplying both numerator and do*
nominator of —
2. The fraction f by 2.
8. The fraction i by 9.
4. The fraction ^jy by 10.
5. The fraction 11 by 7.
6. The fraction ^ by 3.
7. The fraction -/^ by 8.
8. The fraction ^^ by 9.
9. The fraction Jf by 4.
10. Explain the effect of dividing both numerator and de-
nominator of the fraction ^ by 3.
Answer. — Dividing both namerator and denominator of the fraction
A by 3 gives i for a result, which expresses i as many parts, each
part 3 times as largo as before. Therefore, the value of the fraction is
unaltered, and -^ =' i>
Explain the effect of dividing both numerator and denomi-
nator of —
11. The fraction Jf by 8.
12. The fraction 1^ by 9.
13. The fraction ^ by 8.
14. The fraction ^ by 18.
15. The fraction ^^ by 7.
16. The fraction ^^ by 41.
17. The fraction f ^ by 5.
18. The fraction /y8y by 39.
139« Lowest Terms.
(a.) The numerator and denominator are called terms of
the fraction.
(b.) A fraction is said to be reduced to its lowest terms
when its numerator and denominator are the smallest entire
numbers which will express its value.
(c.) From the preceding explanations, it follows that, -—
1. A fraction may be reduced to lower terms by dividing
both numerator and denominator by any number which will
divide both without a remainder.
2. A fraction may be reduced to its lowest terms by divid-
ing both numerator and denominator by any number which
16
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182 FRACTIONS.
Will divide both without a remainder; then dividing this
result in the same way, and so on, continuing the process till a
fraction is obtained, the terms of which are prime to each
other ; or by dividing both numerator and denominator by
their greatest common divisor.
3. A fraction will always be reduced to its lowest terms
when there is no number greater than 1 which will divide
both its numerator and denominator without a remainder.
1. Reduce -^ to its lowest terms.
Solution. — Dividing both numerator and denominator by 4, their
greatest common divisor, gives §, which expresses ^ as many parts,
each part 4 times as large as before. Hence, xV = $•
2. Reduce f JJif to its lowest terms.
Solution. — Observing (104, II.) that both numerator and denomi-
nator are divisible by 4, we first divide by 4, which gives T^f^lfj or ^
as many parts, each 4 times as large as before.
Observing (104, IV.) that both numerator and denominator of the
last fraction are divisible by 9, we divide by 9, which gives i-Jf^, or j-
as many parts, each 9 times as large as before.
Observing (104, V.) that both numerator and denominator of the
lost fraction are divisible by 11, we divide by 11, which gives t^sVj o'
1*1 as many parts, each 1 1 times as large as before. As the numerator
and denominator of the last fraction are prime to each other, the reduc-
tion can be carried no farther, and f Sji^f reduced to its lowest terms
equals -^sV*
Second Solution. — The greatest common divisor of 38412 and 50293
found by one of the methods of Section IX.) is 396 ; and dividing both
numerator and denominator by it, gives i^^V* or ij^^ as many parts,
each part 396 times as large as before. Hence, f § JH = 1^T«
Note. — The mechanical process by the first solution is merely to
divide both terms, first by 4, then by 9, then by 1 1 ; while by the last it
is to find the G. C. D. of both terms, and divide them by it. The first
method will usually be the most convenient, when the divisors can be
readily perceived.
(d.) The pupil should be careful not to decide that any
fraction is incapable of reduction till he has carefully tested
it by some of the processes of Section IX.
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FRACTIONS.
189
Reduce each of the following fractions to its lowest terms.
8. U-
4. if.
«. Hi-
6. «.
7.
8.
9.
10.
If
a-
Hi-
ll.
12.
13.
14.
140. Cancellation.
(a.) When, as is sometimes the case, the factors of the
numerator and denominator are given, labor may be saved by
reducing the fraction to its lowest terms before multiplying
the factors together.
(b.) In writing the work., it is well to draw a line through
the factors which have been divided, and to place the quotients
above those in the numerator, and beneath those in the de-
nominator.
(c.) The numbers by which we divide are said to be can^
celled, and the process is called cancellation ; but it is identi-
cal in principle with other cases of reducing fractions to theii
lowest terms.
1. Reduce
8 X 15
9 X 16
to its lowest terms.
Solution. — Cancelling 8 from the factors 8 of the numerator and 16
of the denominator, (i. e., dividing each by 8,) gives 1 in place of 8,
and 2 in place of 16, and makes the fraction express i as many parts,
each 8 times as large as before.f
Cancelling 3 from the factors 15 of the numerator and 9 of the de-
nominator, gives 5 in place of 15, and 3 in place of 9, and makes the
fraction express -^ as many parts, each 3 times as large as before.^ As
no farther division can be made, the remaining factors are to be multi-
plied together, which gives | for a result
• Solution. 64 and 81 are prime to each other, and hence f4 cannot b*
reduced to lower terms.
t For multiplying by |- of a number gives -J as large a product as mul-
tiplying by the number would give.
X For multiplying by -^ of a number gives -J- as large a product as miil«
tiplying by the number would give.
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18< FKA.CnON8.
TIm work woold be written thus : —
1 5
$ X U _ 5
8 2
2. Reduce 3- x 12 X 4 X 11 X 21 *°'^^*>^^''*^"°*-
Solution. — Cancelling the factor 12 from numerator and denomina-
tor, gives 1 in place of each. Cancelling 7 from the namerator and
from the 35 in the denominator, gives 1 in place of the former, and 5 is
place of the latter. Cancelling 5 from the denominator and from the
25 in the numerator, gives 1 in place of the former, and 5 in place of
the latter. Cancelling 4 from the denominator and from the 36 in the
namerator, gives 1 in place of the former, and 9 in place of the latter.
Cancelling 1 1 from the namerator and denominator, gives 1 in place of
each. Cancelling 3 from the 9 in the namerator and from the 21 in the
denominator, gives 3 in place of the former, and 7 in place of the latter.
As no further redaction can he made, we multiply the remaining factors
together, which gives -^ = 2^.
The written work would bo thus : — .
115 1
/g X ar X ^$ X 0$ X ii _ 1^ _ oi
ii$ X i^ X 4 X ii X fii^ 7 '^ 7
^1117
1
Or, by omitting to write the factors which are equal to 1, as we may
do without ambiguity, we have the following more convenient form.
3
5
a X Tt X fi$ X 0$ X a _ 15 _ 1
fi$ X Xii X i X tX XiH" 1 ^ 7
^
8. Reduce
5 X 7 X 11 X 12 X 15 X 18 X 2
4X5X3x11X7X36X5X6
Its lowest terms.
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FRACTIONS. 18S
4 X $ X X i:i X "H x00x«Jx0~2
<i 2
. ^ , 48 X 30 X 49 X 64 X 27 ^ . ,
^ ^"^"*^ 21 X 82 X 36 X 42 *° "^ ^°^«*
terms.
iifinoer. — 3
4 10 jr )g
^0 X 00 X <r0 X 0^ X jgt _ 120 _ ojj
iix%ixUx 4^ ~ 1 ~
3f <J
Reduce each of the following to its lowest terms.
7 X16X18X5X 9
5.
6.
7.
10.
11.
12.
18.
20 X 14 X 9 X 9 X 11
8 X 86 X 28 X 48
24 X 72 X 18 X 15
6X7X8X 9 XIO X 11 X 12
7 X 8 X 9 X 10 X 11 X 12 X 18
96 X 65 X 8 5
25 X 91 X 48
14 X 5 X 9 X 80 X 16 X 17
8 X 17 X 14 X 27 X 11 X 8
72 X 49 X 81 X 38
77 X 84 X 6 X 27 X 99
14 4 X 108 X 625 X 121
875 X 64 X 99 X 847
815 X 148 X 64 X 221
119 X 169 X 209 X 125
1881 X 848 X 6859
1468 X 2527 X 847
16*
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16C FRACTIONS.
,. 4959 X 3487 X 2491
6061 X 53 X 2853 X 47
141. Compound Fraction.
(a.) A coBfPOUND FRACTION IS a fraction of a fraction
as f of I, § of t of A-
{h.) A compound firaction is equivalent to a fraction mul«
dplied by a fraction.
Thus : i of f s= i times f ; f of ^ of A = ? times ^ times
^. (See 133.)
(e.) In a componnd fraction, the yalne expressed hj one fraction is
made the unit of another. J of ^ means f of the quantity ^, or 3 such
parts as would be obtained by dividin;^ -f into 4 equal parts. § of | of
•/5r means § of the quantity ^ of t^^, and ^ of -fxs means f of the
quantity -f^,
(d.) In reducing compound fractions to simple ones, it is important
to notice which fraction is made the unit of the other, as that is the one
on which the operation is to be performed.
1. Whatisf of f}H?
Solution, f of H4i = 8 times i of f J JJ ; i of fJHi fonnd by
dividing 5747 by 9, is ^f Jf », and 8 times this result, found by multiply-
ing 638f by 8, is — JL Hence the following forms of written work.
9436
First Farm, Second Form.
a = 5747 9 ) 5747
i of a = b = (
8 times b s siOSf == f of a. 638^
Hence, S of JiH = WJI^ ®
5108|
Hence,! of HH«=ti*l*
2. What is tSt of fit ? 5. What is | of ^^f ?
3. Whatis^of T^ft^? 6. Whatis^Jof |H*f?
4. What is ^^ of |5f f ? 7. What is ff of tl?f ?
8. What is i of 437f ?
Solution, i of 437^ = 54, with a remainder of 5f , which, reduced
tc seventha. « V^; ^ of ^yJ^ = f . Hence, i of 4S7f »« 54f .
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FRACTIONS. 187
NoTB. — Compare tho abore with ** What is 7 of 437 weeks and 5
days?"
9. What is i of 8539^ ?
10. What is f of 827f ?
11. Whatisf of 5827J?
12. What is t of 27981 ?
13. What is Jof 1397^?
14. What is I of 4355^^1?
Second Method of deduction.
(e.) When the numerator of the unit of the compoiind
fraction is not a multiple of the denominator of the other
fraction, the preceding solution will give a fraction in the
numerator of the result. In all such cases the following solu-
tion should be adopted. Indeed, applying, as it does, tho
principles of cancellation, it really includes the preceding.
15. Reduce J of |^ to a simple fraction.
First Solution. — Since f of any fraction must eqnal 3 times as many
parts, each ^ as large as before, f of |- may be found by multiplying
the denominator of f by 8, and the numerator by 3. This gives the
following written work.
3 ,4 _ 4x $ _ 1_
8°9 0Xg"~6
3 2
Second Solution, i of i may be expressed by making 8 a factor of
the denominator, and v of f- must be 3 times this result, which may ba
expressed by making 3 a factor of the numerator. This gives the same
written work as before.
Note. — In writing the work of such examples as the above, the
fraction on which the operation is to be performed should always be
written first
16. Reduce f of ^f of 3^ to a simple fraction.
F^rst Solution, 3-ft-, or ^, being the number on which the opera-
tion is to be performed, should be written first. J|*of this most equal
15 times as many parts, each J^ as large, and may be expressed by
making 15 a factor of the numerator, and 28 a factor of the denominator.
I of this must equal 8 times as many parts, each | as large, and hence
may be expressed by making 8 a factor of the numerator, and 9 a (Mot
of the denominator. This would give the following work.
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18S FRACTION^.
5 5 2
* f M f si - g^ X a'j? X _ 50 _ .17
9 ° 28 ° 11 ~ 11 X ;20 X ~ 33 - 33
<< 3
Second Solution, S-ft-, or f^, is the number on which the operatioQ
is to be perfonned, and shonld therefore be written first. ^V ^^ ?f
maj be expressed by making 28 a factor of the denominator, and j^|
of -H n^QBt be 15 times this result, which maj be expressed by making
15 a &ctor of the numerator, i of this result may be expressed by
making 9 a fkctor of the denominator, and f , or 8 times the last result,
by making 8 a factor of the numerator. This would give the same
written woHl as before.
KoTB.— Itwill be seen that both of the above solutions give the
same numerical process, viz., to make all the numerators of the com-
pound fraction factors of the new numerator, and all the denominators
factors of the new denominator, and then cancel and reduce.
Reduce each of the following fractions to simple ones.
22. x^ofAoflSf
23. f of 4 of 3}.
24 ^of Aof J*.
25. II of H of ^}.
17. foff^.
18. if of if..
19. iofif-
20. fof^ofl.
21. i off of if.
26. iof iof f of iof if of 5f
27. Aof f of 7»^of Aof 9*.
28. fof A of ft of f J of If
29. f of J of ft of ff of 13^.
30. f of t of f of I of A of fi of if of f I of 62f .
143. MuUipUcation of Fractions.
1. What is the product of f X f ?
Sotution. i multiplied by f = | of t> which, found by multiplying
the numerator of f by 3 and the denominator by 8, gives the following
written work.
* X ?
9^8
4X0
~ X
3 2
1
— 6
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FRACTIONS.
m
(a.) By reading the sign X as timesy we have —
f times i B f of f , which, found by multiplying the nnmecmtor ^
} bj 4 and the denominator by 9, gives the following written work.
9^8-
0X4 1
0X ~ 6
2 3
What 18 the product of —
2. fXiJ?
6. fXfXS?
6. 4iX4?X8J?
7. 3iX6iX A?
8. 13f X 23fr X 8^?
9. 5iXJof5A?
10. fxfxiixii?
(5.) The following form of explanation may be adopted
when a very thorough analysis is required.
11. What is the product of ^ X f i ?
Solution, — Pirst write ff as the number on which the operation
is to be performed ; we then have f f multiplied by 1 eqnals f f , and
mnltiplied by pV ^i^ equal ^ of this result, which may be expressed
by making 98 a factor of the denominator. If multiplying by ^ gives
this result, multiplying by f ^ roust give 81 times this result, which may
be expressed by making 81a factor of the numerator. Hence, —
49
81 _ X fii
98
04 X
2
2
'SoTB. — It will be seen that all the forms of solntion give eimilar
forms of written work, the numerators of the several fractions being in
all cases factors of the numerator of the product, and the denominator!
factors of the denominator.
What is the product of —
12. fX/jXH?
13. MX??x+H?
14. ax^xa?
15. T»A X WnnX-fil^?
16. .83 X .007 X .49?*
17. 8.7 X .43 X .006?*
• For models of written work, see ITlst page, solution to 8th example,
and 17!id psge, Nota.
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190 FRACTIONS.
18- IttXHXH? I 20. .824X4.3?
19. 43.79X25.7? I 21. 6750 X .6750?
22. 5.06 X 300 X. 2 X. 0004?
23. .4 X .3 X .2 X .6 X .2 ?
24. 4.974 X 1.0007 ?
143* Reduction of a Vulgar Fraction to a Decimal Form
1. Reduce § to a decimal fraction.
JSolution, i, or J of 1, = i of 7 = J of 70 tenths = 8 tenths, with
a remainder of 6 tenths. But 6 tenths =: 60 hundredths, and ^ of 60
hundredths s= 7 hundredths, with a remainder of 4 hundredths. Bat 4
hundredths »= 40 thousandths, and i of 40 thousandths = 5 thou-
sandths. Hence, J = .8 + .07 + .005 = .875
The work may be written thus : —
8)7000 Proof. .875 = ^ft^V = f
.875
2. Keduce |g to a decimal fraction.
Solution, f I, or §| of 1, = sV of 25 = yV of 250 tenths = 8
tenths, with a remainder of 26 tenths. But 26 tenths = 260 hundredths,
and yV of 260 hundredths «= ; &c Hence the following written woHe.
28 ) 25.00 .892857 A = ^«»^ or dropping the ^Vt
2.60 or ^ of a millionth, we shall have the
■ ■ approximate value of ft = .892857
.0240
.00160
.000200
.000004
Proof. .892857 J « 892857+ times looio oo = A^i^filUl ^
<ciF<joTrini '^ ft*
Note. — Compare the abore solutions with 87, (c.)
(h.) The fractions may also be reduced by the following
0olutions : —
7 X 10
J, or i units, = 10 times as many tenths = — g — tenths «=» 10
as many hundredths as tenths = 7 X 10 X 10 hundredths ■» 10
8
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FRACTIONS. 191
timei as many thousandths as hundredths s= 7 X 10 X 10 X 10 jj^^^
8
sandths.
By cancelling the last, we have, —
5 5 5
g g ■ — • thousandths = .875
i
NoTB. — The above is eqaivalent to the following : —
5 5 5
7 _ T Xi0 X i0 X i0 _ 875 _
8 X 10 X 10 X 10 ~ 1000 — ^^
i
i
25 25 25 X 10
So, — , or — units, = 10 times as many tenths = — ^ — teaHif,
B 10 times as many hundredths as tenths = 5 X 10 X 10 j^^^^
diedths,&c
5 5
TT«n«. 25 _ 25 X 3^0 X a:0 X 10 X 10 X 1 X 10
^28- M ~
U
7
6250000
millionths = — = millionths = .892857^
NoiB. — The above is equivalent to
5 5
25^ _ 25 X a:0 X 1^0 X 10 X 10 X 10 X 10 _
28 "~ ^0 X 10 X 10 X 10 X 10 X 10 X 10 ""
U
7
6250000 6250000 , ^^„^„, „«„„„.
700000C = -T— «^ '^^^^^ = •«»2857+
(e.) It is obvious that ff cannot be reduced to an ezact^
equivalent decimal form, for introducing into the numerator
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m FRACTIONS.
any number of factors each equal to 10 will not enable us to
cancel the factor 7 from the denominator. As the same prin-
ciples apply to any other fraction, it follows that, —
(d,) No vulgar fraction can he reduced to an exactly equiv*
alent decimal form^ if^ when redticed to its lowest termSy its^
denominator contains other factors than such as are found in
10, L e., other than 2'« or 5*s; and conversely, that, —
(e.) Every vulgar fraction which, when reduced to its loW'
est terms, contains in its denominator only such factors as are
found in 10, can be reduced to an exactly equivalent decimal
form, and wiU contain as many decimal places as there are
2'« or 5's to be cancelled from the denominator.
Illustrations. ~ s=s •-^t and hence can be reduced to an exactly
equivalent dedmaL MoreoTcr, it wiU contain four decimal places ; for
10 mast be introduced 4 times as a factor into the numerator, to cancel
8* fiom the denominator.
IS 13
Aeain. r-- aes ■ ^^ . and hence can be reduced to an exactly
® 250 2 X 5* "^
equiyalent decimaL Moreover, it will contain 3 decimal places, for 10
must be introduced 3 times as a factor in the numerator, to cancel the
foctors of the denominator.
27 27
Again, gr =* ^2 y lo * *"^d hence cannot be reduced to an exactly
equivalent decimal form.
(/.) Vulgar fractions which cannot be exactly reduced
give rise to BEPEATma or oiRCULATiNa decimals.
Beduce each of the following to a decimal form, carrying
the division, when only approximate values can be obtainedi
to six decimal places.
8. m-
7. A.
11. ff.
*- u-
8. if-
12. If
6. a.
9. tft.
13. if.
6. if. J
10. f
14. ii-
14L4L» Fractional Parts of Denominate Numbers.
(a.) What is the value of f of a mile in whole numbers ot
lower denominations, i. e., in furlongs, reds, yards, &c ?
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FKACTIONS. 191
Firtt Solution. — This example may be solved by the common pro-
cess of compound division, thus : |- of a mile = ^ of 8 miles =
niiles, with 8 miles remaining. But 8 miles = 64 fdrlongs, and i of 64
furlongs == 7 furlongs, with, &c
Second ^Solution. — Since 1 m. = 8 fur., |- of a mile must equal j
of 8 fur., which is li fur. Since 1 fur. = 40 rods, ^ of a fur. must
equal i of 40 rods, which is 4^ rods. Since 1 rod = 5^, or V* yd8-> i
of a rod must equal i o{ -^ yds., which is 2|^ yds. Since 1 yd. aa
3 ft, J of a yd. must equal J- of 3 ft, which is 1^ ft. Since 1 f t «=»
12 in., -J of a ft. must equal i of 12 in., which is 4 in. Therefore, f
of a mile = 7 fur., 4 rd., 2 yd., 1 ft., 4 in. The work may be written
thus:--.
m. fur. for. fur. rd. rd.
8 8 X 8 64 .1 1 I X 40 40 4
rd. yd. yd. yd. ft. ft.
4 __ 4 X 11 _ 22 4 4 ^ 4X3 ^4 1
9 9X2 9 * ^^ 9 9 '^ 9 "" d' ^'^ 3
ft. in.
_ s- as 4 in. Hence, # m. = 7 fur. 4rd.2yd.lft.4in.
8 3 "^
Third Solution. — Since there are 8 furlongs for every mile, there
must be 8 times as large a part of a furlong as of a mile, or, in this in-
stance, 8 times f of a furlong, which is 7 j- furlongs. Since there are 40
rods for every furlong, there must be 40 times as large a part of a rod
as of a furlong, or, in this instance, 40 times i}- of a rod, which is 4|-
rods. Since for every rod there are 5^, or -y- yards, there must be -^
as large a part of a yard as of a rod, or, in this instance, -V* of |- yard,
which is, &c. The written work would bo the same as before.
(6.) The only difference between the processes of this and the pro-
Tious article is the difference between decimal and denominr.te numbers.
In the former, a unit of any denomination equals 10 of the next lower,
while in the latter the number of units of a lower denomination to
which any unit is equal, varies with the denomination of the unit con-
sidered.
What is the value of each of the following in whole num*
bers of lower denominations ?
1.
^ of a £.
6.
•]Bf of an acre.
2.
ii of a gal.
7.
\^ of a nule.
8.
^1 of a bu.
8.
-^ of a ton.
4.
tl of a week.
9.
;ft of a lb. T.
&
iJ of » Cd. ft.
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194 FRACXION8.
(c.) Reduce .7925 of a £ to whole numbers of lower d6>
nominations.
Firti Solution. — Since £1 = 20 s., .7925 of a £ must equal .7925
of 20 8. =3 20 times .7925 s. = 15.8500 s. Since 1 s. = 12 d., .85 of a
t. most equal .85 of 12 d. s= 12 times .85 d. s= 10.20 d. Since 1 d. *•»
4 qr., .2 of a penny must equal .2 of 4 qr. =s .8 qr. Hence, .7925 of a
< ca 15 8., 10 d., .8 qr.
.7925 £.
20
WBITTBN WOBK.
Or, byo
tipliers, we
mitting to write the mul
have, —
15.8500 8.
.7925 £.
12
15.8500 8.
10.20 d.
4
10.20 d.
.8 qr.
.80
Hence, .7925 of a £ = 15 s. 10 d. 8 qr.
Seeond Solution, — Since there are 20 s. for every pound, there must
be 20 times as large a part of a shilling as of a pound, or, in this case,
SO times .7925 s. &= 15.85 s. Since there are 12 d. for every shilling,
there must be 12 times as large a part of a penny as of a shilling,
or, in this case, 12 times .85 s., &c. The written work is the same as in
the last solution.
KoTB. — The student should be careful to multiply only the fraction*
a1 part of each number.
What is the value of —
1. ^45 of a ton?
5.
.2876 of a gal.?
2. .6875 of a mfle ?
6.
.625 of a bu. ?
8. .8426 of a lb.?
7.
.817of acu.yd.?
4. .754ofa£.?
8.
.2569 of a cwt ?
9. What part of 1 mile is 7 fur. 4 rd. 2 yd. 1 fb. 4 in. ?
Solution, — Since 1 in. == i^ of a foot, 4 in. must equal -^j or i of
m foot, to which adding the 1 foot gives \i ft. &== f of a foot. Since
1 ft. = i of a yd^ i of a foot must equal J of i of a yd., or f of a yd.,
to which adding the 2 yards gives 2| yd., or ^ yd. Since I yd. «
A^ of a rd., V of a yd. must equal ^ of -ft- of a rd., or J of a rd., to
Which addmg the 4 rd. gives, &c.
When the work is written, the following form may be adopted :^
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FRACTIONS. 19ft
It. ft. yd
yd. rd. rd. flir.
2
4 22 ^SflL _ 2 4 4 40 "til ^ 1 1
ftir. m.
8
.^1 64 "^64^,1 8
Hence, 7 fur. 4 rd. 2 yd. 1 ft. 4 in. = f of a mUe.
Proof, — Reduce f of a mile to fur., rd., &c.
Serxmd Solution, — Since there is xV of a foot for each^inch, theri
must be iV <u many feet as inches, or, in this case, iV of 4 ft. s= iV« or
-^ of a ft. Since there is -^ of a yard for each foot, there must be | a«
many yards as feet, or, in this case, ^ of '•J, or i yd. =, &c
The written work is the same as in the last solution.
Note. — Tlie method here given is usually preferable to that of
iSOy solution of 72d example, inasmuch as it keeps all the fnictioDf
reduced to their lowest terms, and enables us to perform very many, If
not most, such examples without writing any figures.
What part —
10. Of 1 A. is 2 R. 36 sq. rd. 8 pq. yd. 2 sq. tt. 36 sq. in. T
11. Of 1 gal. is 3 qt. 1 pt. 1» gi. ?
12. Of 1 Cd. ft. is 10 cu. ft. 1382§ cu. in. ?
13. Of 1 T. is 17 cwt. 3 qr. 2 lb. 12 oz. 7^ dr.
14. Of 1 lb. is 6 oz. 13 dwt. 8 gr. ?
15. Of 1£ is 9 8. 5d. 1-J qr?
16. Of 1 circumference is 155° 4' 36 [|"?
17. Of 1 lbis5S 6309 6f gr.?
18. Of 1 w. is 3 da. 10 h. 17 m. 84 sec ?
(a.) Should it be required to give the answers to suoh
questions as the above in a decimal form, it will only be
necessary to reduce the vulgar fractions obtained by the pre.
ceding process to equivalent decimals.
(c.) The following process may also be applied : —
19. What part of a pound is 13 s. 7 d. 2 qr. ?
Solution, — Since for every farthing there is }^ of a penny, there mar
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196 FRACTIONS.
be ^ as maoj ponce as farthings, or, in this case, j- of 2 d. =» 5 of a
]>enny, to 'iirhich adding the 7 d. gives 7.5 d. Since for every penny
there is ^ o{ a shilling, there mnst be iV «^ many shillings as pence,
or, in this case, ^V of 7.5 s. =, &c.
The most convenient form of writing the work is to arrange the
■ambers expressing the varioas denominations in a vertical colomu,
thM: —
4 ) 2.0 qr.
12 ) 7.500 d. = 7 d. 2 qr.
20 ) 13.6250 s. = 13 s. 7 d. 2 qr.
.68125 £ = 13 8. 7 d. 2 qr. = Am,
In like manner perform the following : —
20. What part of 1 bu. is 5 pk. 3 qt. 1 pt. ?
21. What part of 1 lb. is oz. 13 dwt. 8 gr. ?
22. What part of 1 gal. is 3 qt. 1 pt 3 gi. ?
28. What part of 1 Cd. fc. is 10 cu. ft. 1382f cu. in. f
149* To find a Number from a Fractional Part of iU
1. 5861 = ^ of what number ?
First Solution, 6861 = ^ of 7 times 5861, which is 41027.
Second Solution, If 5861 = f of some number, -J, or the number
Itself, must equal 7 times 5861, which is 41027.
Proof, i of 41027 = 5861.
Of what number —
2. Does 3498 = i?
8. Does 59387 = ^ ?
4. DoesitJ = -J?
8. 3476 = f of what number?
Solution, — If 3476 = f of some number, y of that number must be
i of 3476, which is "^/-^, and f , or the number, must be 9 times thia
result, and may be expressed by multiplying the numerator by 9.
Hence we have the following written work : —
869
»AH^ 8 ,UHX9 8' 7821 8 , „^,^,
S476 = 5 of g— = - of -g- = 5 of 3910J
2
s: Answer.
, Proof I of 3910 i =s 3476.
5. DoesfH = i?
6. Does 68.46 = .1 ?
7. Does 327.93 = .0001 ?
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FRACTIONS. IW
9. If = Jl of what number ?
Solution, — If ^ = f I of some number, ^V of that number must
equal ^ of ff, which may be expressed by making 35 a factor of the
denominator, and f f , or the number, must be 36 times the last product,
whirih may be expressed by making 36 a factor of the numerator
This gives the following written work : —
7 2
4^ X U _ 14 _ ^
$4^rT$ - IB "" ^'*'-
3 6
Proof, — See if t| of {\ is equal to Jf .
Of what number —
10. Does 5496 = f?
11. Does 23584 = $f?
12. Does 16875 z= II?
13. Doesi = f?
14. Does A = A?
15. DoesjfL=}|?
16. 8372 = .07 of what number?
Solution, — Since 8372 = .07 of some number, .01 of that number
must equal ^ of 8372, which is 1196, and \%%, or the number, must
equal 100 times this result, which, found by removing the point two
places towards tlie right, is 119600.
Of what number —
17. Does 5987 = .9 ? I 19. Does 79.84 = .004?
18. Does 2.475 = .03 ? 20. Does .58674 = .OOlli
14:6* Practical Problems,
1. K § of a yard of cloth cost $3.74, how much will 1
yard cost ?
Solution, — If f of a yard of cloth cost $3.74, ^ of a yard will cofi
i of $3.74, which is $1.87, and f , or a yard, will cost 3 times $1.87,
which is $5.61 = Ans,
The work may be written in either of the following forms : —
First Form,
1.87
&^^ .= 05.61
17*
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198 FKACTIONS.
Second /Vrm.
2 ; $3.74
$1.87
3
$5.61
Proof — I^ 1 yard of cloth costs $5.61, i of a yard wUI cost J of
$5.61, which is 91.87, and $ of a yard will cost two times $1 87, which
is $3.74.
2. K J of an acre of land cost $8.54, what will an acre cost f
3. If J of a vessel cost $17645, what will the vessel cost ?
4. If 1^ of a cask of oil is 143 gallons, how many gallons
are there in the cask ?
5. If .09 of a lot is worth $594.72, how much is the lot
worth?
6. If .13 of a cargo of coal is worth $213.46, how much
is the cargo worth ?
7. If 4 of a pound of tea is worth f ^ of a dollar, what is
a pound of tea worth ?
Solution. — Since the answer is to be in dollars, we write ff of a
dollar as the number to be operated npon. If ^ of a pound of tea cost
so much, ^ of a pound will cost -J of this, which may be expressed by
making 5 a factor of the denominator ; and f , or' a pound, will cost 7
times this result, which may be expressed by making 7 a factor of the
numerator.
This gives the following written work . —
5
Ji$ X 7 ^35 ^, 8
'27 X $" '"'27 - ^"27 - ^
8. If f I of a bale of cotton weighs if of a ton, how
much will the bale weigh ?
9. Bought §J of a gaUon of oil for f J of a dollar. How
much would a gallon have cost at the same rate ?
10. If I of a yard of silk cost $1,572, what will f^ of a
yard cost ?
Solution. — Since the answer is to be in dollars, we first write $1,572
as the number to be operated upon. If | of a yard cost $1,572, ^ of a
yw4 will cost ^ of $1,572. which may be expressed by writing 8 undei
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FRACTIONS. 199
f 1 572 as a denominator, and f , or a yard, will cost 9 times this result,
expressed by making 9 a factor of the numerator. If 1 yard costs so
much, f ^ of a yard will cost f ^ of this, found by making 20 a factor
of the numerator, and 27 a factor of the denominator. This gives the
foUowina: written work : —
.131
.^0{{ 10
Note.— Questions like the above can always be resolved mto two
or more simple ones. Thus : If f of a yard of silk costs $1,572, what
will 1 yard cost ? If 1 yard costs the last result, what will f f^ of a
yard cost 1
11. How much will ^ of a cord of wood cost, if ^ of a
cord cost $3.85 ?
12. If a ship sails 157 miles in -/j of a day, how many
miles will she sail in 8| days ?
13. A man sold 8^^ tons of iron for $384, and afterwards
sold 9§ tons at the same rate. How much did he receive for
the last lot ?
Suggestion. — Reduce the mixed numbers to improper fractions.
The question then will read, — A man sold -^7^ of a ton of iron for
$384, and afterwards sold ^1^ of a ton, &c.
14. How much must be paid for 5^ acres of land, when
$868 are paid for 2§ acres ?
15. If 6| barrels of flour cost $38.25, how much will 14|
barrels cost ?
16. If I of a dollar will purchase 7-^ lb. of coffee, how
many pounds can be purchased for $5^ ?
Solution. — Since the answer is to be in pounds, wo write 7 i^fi '-*
f ^, as the number to be operated on. If f of a dollar will purchase f ?
lb. of coffee, i" of a dollar will purchase ^J- of ^f lb., and f , or a dollar,
will purchase 8 times this result, and $5|-, or $^, will purchase -^^ of
Ihis last result. The work, then, would be written thus : —
2 3
80 X0X)gjr_48O 7
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200 FRACTIONS.
17. If 12f yards of calico are given for 17j^ jards of
sheeting, how many yards of calico must be given for 25jSg
yards of sheeting ?
Sugfjestion, — Observe that as the answer is to be yards ofoalioo^ tho
nmnber of yards of calico is that to be operated on.
18. If \^^ tons of hay cost $22 J, how much will 9| tons
cost?
19. If 13^^ lb. coffee are worth as much as 3f lb. of tea,
how many pounds of coffee are worth as much as 7^ lb. of
tea?
20. If 8J yards of silk are worth as much as 24J yards
of gingham, how many yards of gingham can be obtained for
67+ yards of silk ?
21. When 3| yards of silk, worth $1^ per yard, are given
for If yards of broadcloth, how many dollars should be given
for 1 Yi yards of broadcloth ?
14:7. Division by Fractions.
(a.) Since 1=J=|=J=:|, &c., it follows that there
must be two times as many halves, three times as many
thirds, four times as many fourths, &c., as there are ones in
any number.
Or, which is the same thing, —
(h.) Since l = 2Xi, = 3Xi, = 4Xi, &c., it follows
that there must be twice as many times ^, three times as
many times «J, four times as many times i, five times as
many times ■^, &c., as there are times 1 in any number.
(c.) But the quotient of a number divided by 1 equals the
-number itself; hence, the quotient of a number divided by ^
must equal twice the number ; divided by ^ must equal tllree
times the number ; by ^, four times the number ; &c
Note. — This is bat an application of the principle, that if on«
Dnniber contains another a certain number of times, it will contain half
iliat number twice as many times ; ^ of it three times es many times
i of it four times as many times ; &c.
For example, —
84 -h 12 = 2 ; 24 -^ ^ of 12, or 6,= 2 X 2, or 4.
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FRACTIONS.
Ml
24 -I- ^ of 12, or 4, = 3 X 2, or 6 ;
24 -*- i of 12, or 3, = 4 X 2, or 8.
In like manner, —
1 -j-l = 1 ; 1 -^ i of 1, or ^, = 2 X 1, or 2.
1 -f- 4" of 1, or -J-, = 3 times 1, or 3 j
1 -5- i of 1, or i, = 4 X 1, or 4.
In like manner, —
7 -*- 1 = 7 ; 7 -5- i of 1, or i, = 2 X 7, or 14.
7-f-i of 1, or i, =5 X 7, or 35;
7 -h i of 1, or ^, = 9 X 7, or 63.
1 . What is the quotient of 9 -r J ? ♦
First Solution, d divided by 1 = 9 , hence, divided by i, it must
equal 4 times 9, or 36.
Second Solution, Since 9 contains 1, 9 times, it must contain ^, 4
times 9 times, dr 36 times.
What is the quotient —
2. Of4-^i?*
7.
Of 4.736 -^ .01 ?
3. Of 311 ~i?
8.
Of .75 ~ .0001 ?
4. Of287-T-i?
9.
Of 9000. -f .01 ?
5. Of 347 -7- tV?
10.
Of .0007 -T- .001 ?
6. Of 347^.1?
11.
Of 864 -T- .0001 ?
12. What is the quotient of | -7- ^ ?
First Solution, |- divided by 1 = J, and divided by i must equal 6
times J, which, by cancelling, = 3 times J = ■^, = 5^-.
Second Solution. Since ^ contains 1, f times, it must contain ^, 6
times i times, = 3 times J = '^- = 5^ times.
Bv either form of solution, the work may be written thus : —
'5' . 1
7
X
21
^1
8 • 6~
4
4 "
--h
What is
the quotient
—
13.
Of
t-Hi?
16.
Of
«-h
iV?
14.
Of
IH-tV?
17.
Of
m^
-tV?
15.
Of
ii-^^V^
18.
Of
m\!
-^T^t7?
• These questions are really equivalent to, " 9 = how many fourthB ? "
' 4 c= how many thurds ? " (See 126, 16th Example.^
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202 FRACTIONS.
19. What is the quotient of ^^ — f ?
Firsi Solution, yf divided by 1 = i^, and by ^ mnst oqoal timef
if ; if it contains j so many times, it must contain § only i aa manj
times.* Hence, —
2 3
16 , 8 _ 3^0 X _ 6
21 • 9^ fit X $'^ ^
7
Second Solution. Jf divided by 1 equals ^f , and divided by i must
equal 9 times ^ ; if the quotient by ^ equals so much, the quotient by
f mnst equal i of this result.* Hence, —
2 3
16 . 8 Hx 6 , .
2i--9 = ;2r3r^ = r"^^^^^5:^
7
What is the quotient —
20. Of^-^Vj?
21. Of if -^ it?
22. Of 8ft ~ J?
23. Of 2f| -r^?
24. Of ilf-MH?
29. What is the quotient of 675 -7- .9 ?
The work may be written thus : —
a = 675 = dividend
10 X a = b = 6750 = quotient by .1
i of b = 750 = quotient by .9
What is the quotient —
30. Of 8.47 -T- .07 ? 33. Of .75 -^ .05 ?
25. Of ^f-^ii?
26. Of824-rf?
27. Of 617 -^T^?
28. Of 675 -T- t^?
31. Of 75. -r- .05 ?
32. Of 75. -^. 005?
34. Of .075 -^ .005 ?
35. Of .00084 -^ .0012?
• In accordance with the principle, that if a number contains anothft •
certain number of times, it will contain 8 times that number only |- u
many times ; 12 times the number only xV as many times, &c.
Thus, 72 -r 2 = 36, and 72-7-6 times 2, or 12, = i of 36, or 6.
48 ~ 3 = 16, and 48 ~ 8 times 3, or 24, = i of 16, or 2.
7 ^ J = 56, and 7 -r 5 times i, or f, = -J of 56, or V" = H^.
t-rl^T- V-,andF-f7timesi\:. di -/ly, = r of V, or J§.
t Rcdube to an improper fraction.
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FRACTIONS. 208
14:8. Process of Division generalized,
(a.) Since the quotient of any number divided by -J = 5
times the number ; by ^ = 9 times the number ; by -j^ =
13 times the number, &;c ; and since the quotient of a num-
ber divided by f = J its quotient divided by \ ; divided by
f =: ^ of its quotient divided by ^ ; divided by -}^ = i\- of
its quotient divided by ^, &c, it follows that the quotient of
a number divided by —
^ = 4^ of 5 times the number, = f of the number ;
|. = ^ of 9 times the number, z=: | of the number ;
■J-J = tV ^^ 1^ times the number, = -Jf of the number ;
.07 z= ^ of 100 times the number = -i^^ of the number;
and, universally, that —
The quotient of any number divided hy a fractioHy is equal
to the product of thai number multiplied by the fraction
inverted.
Illustraiions, — 1. To divide by j-, we have only to multiply by 9 and
divide by 8.
2. To divide by ^, we have only to multiply by 17 and divide
by 3.
3. To divide by .03, we have only to multiply by 100 and divide by
3, or, which is the same thing, to divide by 3 and remove the point 3
places towards the right.
4. To divide by .000037, we have only to multiply by 1000000, and
divide by 37, or, which is the same thing, to divide by 37, and remove
the point 6 places towards the right.
1. What is the quotient of -j&i- -r- -f^ ? %
5 , 10 _ f? X 18 _ 18
11 • 13 "" 11 X i0 ~ 22
2. What is the quotient of. 520.6 -i- .011 ?
Am. — The quotient of 520.6 -*- .011 maybe obtained by dividing
620.6 by 11 and removing the point three places towards the right,
thns: —
FirH Form, Second Form,
jQll ) 520.6 Oil ) 520.600
" 47327^ 47827.j^
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20i FRA.CTION9.
NoTB. — The second ibrra of writing the work differs from the flnt
only in this, — that in it as many zeros are annexed to the dividend as
would be necessary were the point actually changed before performing
the division. Great care is necessary, by cither of these forms, to insure
that the point is placed correctly in the quotient ; and if in any case
ther9 is a doubt as to its true position, the work should be written io
fall, as in Uie model given after example 29th, 147*
3. What is the quotient of f of J of ^ -rf of if of Ji ?
Solution. —
g of ^ of j^ ^ g of 25 of ^ = ^5X9X 11"^
5 2
8 X 12 X 21\* 4 X 7l X $ X ^ X i$ Xift
I) - $X
9 X 25 X 22/ '^ $ X ^ X ii X $ X ii X a
10__ 1 ^ ^
9 ■" ^9
Note. — The more ftiU and analytical explanation would be the
following : j®x is the number to be operated upon. J of "ft- may bo
expressed by making 7 a factor of the numerator, and 9 a factor of
the denominator. ^ of this may be expressed by making 4 a factor of
the numerator, and 5 a factor of the denominator. The quotient of
this quantity divided by J^ will be f f of it, and may be expressed by
making 22 a factor of the numerator, and 21 a factor of the denomina-
tor. If ^ is contained so many times, j^f of §-j- must be contained
^ as many times, expressed by making 25 a factor of the numerator,
and 12 a factor of the denominator. If this divisor (jf of §i) is con*
tained so many times, f of it must be contained f- as many times,
expressed by making 9 a factor of the numerator, and 8 a factor of the
denominator. Hence, -—
4,7,8 8, 12, 21
2 5
$ X •;( X 4 X fifi X fi'$ X ^ _io_ ^1
iix^x$xsitxifix$~9~9
* The equation in the parenthesis may be omitted in praotieal operfr*
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FRACTIONS.
205
What is the quotient —
4. Of 13| 4- 8H ? 7. Of .004 -r 25 ?
5. Of llf -r 4^ ? 8. Of .067 ^ .02 ?
6. Of 16f -r 28f ? 9. Of 3287 -r .0004 ?
10. Of f ofiiof |f-^iiof3^^Vof TJ^?
11. Of fof-V^of T^^-^of T^^of 6i?
12. Of i of i of i of i of i of f -T- i of i of tV of A ?
13. Of f of t of f of f -T- /^ of if of li?
14:9* Complex Fractions.
(a.) A COMPLEX FRACTION is One having a fraction in
either numerator or denominator, or in both ; as, — , -i ^
3f' 4? 7|
NoTB. — Complex fractions are usually considered as expressions of
unexecuted divisions, and are read accordingly. Thus, —
(S.) To show their similarity to other fractions, we may
explain them thus : —
_ ss 7 parts of such kind that 3§ of them would equal a unit
(c.) Complex fractions can be reduced to simple fractions
by the ordinary process of division.
54.
1. Beduce —• to a simple fraction.
sa^^ ^=
86* , 87_86 X 4 __ 144
7 '4 7 X 37 259
Beduce each of the following to simple
fractions.
-ft
5. 8*.
12i
8. '^^.
8
^*
•6. a
4^
»•!!•
^^
7. ^^K
Hi
xo. «
* By reducing 6\ to sevenths, and 9^ to fourthf .
18
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f06
FRACTIONS.
{d.) Complex fractions may oflen be reduced to simpls
ones, by reducing them to their lowest terms, —
3 3
Thus : Dividing both terms of —• by I J gives ^ — f. Dividing
both terms of ~ by l5 gives -^ = i-
5
*k'
Beduce each of the followmg in the same manner : —
11. a
5
-k
17. -51
16§
-^•
..a
18. a
--'4
16. ?l.
6i
19. H
7*
(6.) Complex fractions may also be reduced to simple
ones, by multiplying both numerator and denominator by such
a number as will give a whole number in place of each.
20. Reduce — i. to a simple fraction.
Solution, — If 4§ be mnltiplied by 3, or some multiple of 3, and lOi
be multiplied by 2, or some multiple of 2, the result will in each case be
a whole number. Hence, if both terms of the fraction — ^ be multi-
plied by some multiple of both 2 and 3, the resulting fraction will be a
simple one. Multiplying by 6 gives --— = — = -.
In the same way reduce each of the following complex
fractions to simple ones : —
21.
5t
22.
2*
7*
28.
4*
8?
24.
13J
27.
lOf
83i
25.
435f
627£
28.
168g
847""
26.
48*
64f
29.
471
28|
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PRA^CTIONa. 207
IciO. Other Changes in the Terms of a Fraction.
1. Reduce f to an equivalent fraction, having 6 for a
numerator.
Solution, — Observing that the proposed numerator, 6, is two times the
^iven numerator, 3, we multiply both terms by 2, which gives - = 22L?
6 .36
te= -, or we may at once wnte - = -
o 4 8
2. Reduce f to an equivalent fraction, having 10 for its
numerator.
Solutim. — Observing that the proposed numerator, 10, is i?, or -, of
o 4
die given numerator, 3, we multiply both terms by ^ or by 1^, which
«Voo 8 8 X li 10 8 10
gives - =s * = — , or - = — -- .
9 9 X IJ 14' 9 11^
3. Reduce ^ to an equivalent fraction, having 10 for its
numerator.
4. Reduce ^^ to an equivalent fraction, having 9 for its
numerator.
5. Reduce ^ to an equivalent fraction, having 6 for its
numerator.
6. Reduce ^ to twenty-firsts.
Solution, — Observing that the proposed denominator, 21, is three
times the given denominator, 7, we have only to multiply both terms
2 2X3 6
by 3, whicJvgiws - == - =;= --, or, omitting to write the inter-
7 7X3 21
2 6
mediate work, we have - = — .
7 21
Note. — The same result might have been obtained thus : —
21 2 * 2 21 1 21 3
1 = gj", hence - of 1 must equal 7 ^^ ©T *' 7 ^^ 21 " 21' '^^^ *^° *^°^^
q g
rr = rr ; but, in practice, the first form will usually be found most
convenient
7. Reduce f to fifteenths.
Solution, — Observing that the proposed denominator, 15, is — of the
given denominator, 7, we have only to multiply both terms by —> or fii
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?08 FRACTIONS.
8. Reduce | to halves.
Solution, — Observing that the proposed denominator, 2, is - of ihm
given denominator, 9, we have only to multiply both terms bj -
,„^ 5_5Xf^li^ 5 li
i.
9
Reduce —
9. § to ninths.
14
t to thirtieths.
10. ^.j to fourths.
15.
J to twenty-sevenths.
11. T^ to forty-fifths.
16.
^ to ninths.
12. 1 to twelfths.
17.
^ to sixty-fifths.
13. f to forty-ninths.
18.
T^j to twenty-seconds.
191* Reduction to a Common Denominator,
{a,) Fractions having their denominators alike are said to
have a common denominator.
Thus, 4, 3, 6, and X have a common denominator, bat ^ and ^
have not
(5.) In reducing fractions having different denominators to
a common denominator, (i. e., to equivalent ones having the
same denominator,) we first select a convenient number for
the common denominator, and then make the reductions as in
the last article.
. (c.) As far as the denominator is concerned, one number
may as well be selected for a common denominator as another^
but unless the number selected is a common multiple of all
the given denominators, one or more of the resulting numera-
tors will be likely to contain a fraction.*
(rf.) To avoid such an inconvenience, and at the same time
to avoid as far as possible the use of large numbers, it will
usually be best to select the least common multiple of tho
given denominators for a common denominator.
1. Reduce J, ^, |, and ^f to a common denominator.
• For it will be the product of a whole number multiplied by a fra*
ticnal quantity.
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FRACTIONS. 20f
Solution. — We select 72 for the least common denominator, becaaM
it is the least common multiple of the given denominators.
Then, since 72 = 9 times 8, we multiply both terms of the fraction
{ by 9. which gives J- = f f . Since 72 = 6 times 12, we multiply both
terms of the fraction -^ by 6, which gives, &c
Hence, J==?f; ^ = ^5 * = ?f J andif = ^|.
(c.) Many adopt it as a general rule, to select the product
of the given denominators for a common denominator ; but it
usually involves larger numbers than the preceding method,
and is hence much less convenient. The following illus-
trates it.
Solution to preceding Example. — The product of 8, 9, 12, and 24, the
given denominators, is 20736, which we select for the common denomi-
nator. To get this, we multiplied 8, the denominator of i^, by 9 X 12
X 24, and therefore we multiply the numerator, 7, by the same num-
bers, which gives J = "tS^i^- To obtain 20736, we multiplied 12, the
denominator of t^, by 8 X 9 X 24, the product of the other denomi-
nators, and therefore we multiply the numerator, 5, by the same numbers,
which gives -^ 5= AVsfV To obtain 20736, we multiplied 9, the de-
nominator of f, by, &c
(/.) When any of the fractions to be reduced are com-
pound or complex, they must first be reduced to simple ones,
and the simple fractions should be reduced to their lowest
terms, except when to do it would increase the labor of redu-
cing to a common denominator.
(^.) Reduce the fractions in each of the following exam-
ples to a common denominator : —
o 4 2 5 ,7
^- 6' 3' 6'^°^ 15"
Q 3 5 1 1 ,4
tj. __ _, _, _, ana — .
f 9 2 6 21
. 19 23 41 87 , 49
4. — , — , — , , and .
24 36 72 108^ 216
. 24 7 14 37 ,44
^ 25' 9^ 15' 225' ^°^ 76-
18*
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210 FBACTIOICS.
7. .,-,-, and ^.
8. 1?, 15, 15, and i?.
14' 21' 28 25
9 ? of 5, - of A, and ? of 1|.
3 4' 6 15' 9 16
10. ^, L of Ih and L.
^ ' 11 ^ 5i
11. ?i,5,?l,andj
4if ? 9? 9
12. 1 of 15, 1, 5L, 51, and 15.
5 16 32' T^ 7i 32
13. 1, 1?, 15, 5 of 51, ^ and L^l
11 16 22^ 9 2? y^ 100
133. Addition and Subtraction of Fractions.
(a.) Fractions, like other numbers, must be of the sama
denomination, in order to be added or subtracted.
(ft.) To be of the same denomination, they must (121 ,
/8.) be fractions of the same unit, and also have a common
denominator.
^ -I- j- cannot be added in their present form, any more than can 3
pounds and 4 ounces.
■f of a yard and ^ of an inch cannot be added in their present form,
any more than can 3 yards and 3 inches.
(c.) When abstract fractions are given, (i. e., fractions of
the abstract unit, as f , f ,) they may, if simple, be reduced at
once to a common denominator ; but compound and complex
fractions must be reduced to simple ones, and fractional parts
of denominate numbers, as f of a yard, J of a foot, must be
reduced to fractions of the same unit, before they ai*e reduced
to a common denominator.
1. What is the sum ofJ-|-f-|-|-|-fJ + f?
Solution — By reducing to a common denominator, we have —
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FRACTIONS. 211
762n3^2120162218^
8 "^ 6 "*" 3 ■*" 12 "*" i 24 "*" 24 '^ 24 "^ 24 "^ 24
21 + 20 + 16 + 22 + 18 _ 1
24 ^'
2. What is the sum of 87^ + 13^ + 27J + 37^ +
Solution. — The snm of the whole numbers is 212. Hedadng the
fractions to a common denominator, we have i+t + t + A + 'Ar
« if + *» + AS + 4?^ + §* = 2ft, which added to 212 = 214M
> The following arrangement of the nambers shows the resemblance
of the work to compound addition : —
Ones.
36ths.
87i- == 87
18
13f = 13
30
27^ = 27
16
37^ = 37
10
48^2- = 48
21
214 23 == 214||.
3. What is the sum of -^ of a ton -j- f of a quarter ?
Solution. — Bj reducing the -/j of a ton to cwt. and qrs., and adding
the f of a quarter, we should have the following written work : —
iV T. + f qr. = 11 cwt. 2f qr. + i qr. = 11 cwt. 3^ qr. = 11 cwt
3qr. 12 lb. 8oz.
4. What is the value of 343f — 138ff ?
Solution. — By reducing the fractions to a common denominator, wo
have 343| — 138f i = 343|f — 138f f = 204|}.
Or the work may be written thus : —
Ones. 72ds.
343J = 343 16 = Min.
138^ = 138 69 = Sub.
204 19 = 204|| = Rem.
(d.) When several fractions are to be added, it will often-
times be a saving of labor to consider at first only two of
them, and then a third with the sum of these two, and then a
fourth with the sum of these, and so on, till all are added.
Care should be taken to couple tLiem in such a way as to
make the reductions easy.
Thus, in the 24th example below, f + f = 1^ j li + i'^f] taui
2 -h i = 2i = Ant.
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i:i
FRACTIOirS.
Perform the following examples : —
^- 8 + iO-
10.
1
2
1
3*
- \^'.
«• i + l:
11.
3
2
16. i 1.
7^9
4
3
4 5
7 *4. »
^- 8 + 12-
12.
7
7
j7 35 35
8
W
■ 52 "78
8. 4^ + 2f
13.
3i-2f
18. 2** H
9. Sf+lf.
14.
5j_3f ^i i^
19. 8l|__4l5
■
25. 9J — 8f.
17J TI
5 7
- Jl + ^
26. - + 3f + 7i + l.
7 3 5 17
21. 13_1
^^- 9 + 8 + 12 + 36-
18 4
1 49
22 27 9
28. n+-^+m+f^
30 10
5 5 3 2
23. 8* + H
2«- 12 + 6 + i + i-
5 2
^*- 6 + i + 3
+1-
30. 5I + 4A+J3 + I
81. ^+si-^ + 5i-^.
82. llof|of| + 4f+lTit + 7.
83 ^V^^
^^- 24 + 96
4 -.7 4i 2
-ofgof-of-.
357 294 423
^** 853 + 671 + 849-
5''^8-
''^ ^ff^ + Hf
6
6'
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FRACTIONS. 218
10? ,2
6 1
88. = of a bu. + H of a peck.
8 5
89. TT of a m. + t: of a furlong.
H o
17 2
40. — of a £ — Q of a sliilling.
rr Q 1
41. - of a lb. + T of a dwt. -)- - of a grain.
42. g of a bu. + Q of a pk. — — ^ of a quart
5 2 2
43. ^ of a yd. + « of a qr. + ^ of a naiL
O u O
SECTION XL
APPLICATIONS OF FOREGOING PRINCIPLES.
193. Introductory Note,
Ix performing the examples of this section, as of every other, the
stadent should observe that the solutions and models of writing the
work are designed to saggest methods of applying principles, and are
not to be regarded as forms which must be rigidly followed. They
should be deviated from when better forms can be discovered, or will
apply. Indeed, a habit of examining each question carefully before
attempting its numerical solution is one of the most valuable a pupil
can acquire. A distinguished teacher once said, " If I were required,
on peril of my life, to perform a complicated problem in two minuteti I
would spend the first minute in considering how to do it.**
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/
SU FRACTIONS.
ltS4« MisceUaneotAs Problems.
1. How much will H yards of silk cost at $1^ per yard ?
2. If 7} yards of silk cost $14, how much will 1 jard
cost?
8. How many yards of silk, at $1} per yard, can be
bought for $14?
4. How much will a lot of land 28| rods bng and 23|>
rods wide cost, at $2^ per acre ?
5. How many pounds of coffee, at 15f - cents per lb., can
be bought for $8.40 ?
6. How many square feet and inches are there in a board
13^ feet long and 2^^^ feet wide ?
7. How many square feet in the four walls of a room, 15^
feet long, 12^ feet wide, and 8J feet high ?
8. I bought a cask, containing 94^ gallons of oil, at $1,375
per gallon ; ^ of it leaked out, and I sold the remainder at
$1.50 per gallon. How much did I lose by the transaction ?
9. Mr. Whitney is worth $1473.21 more than Mr. Whip-
ple, and Mr. Whipple is worth just J as much as Mr. Whit-
ney. How many dollars is each of them worth ?
Suggestion, — If Mr. Whipple is worth only f- as much as Mr. Whit-
ney, the difference between the values of their estates mast equal ^
of Mr. Whitney's estate.
10. What is the value of | of U of ?f "J" Jf «f.| of
7 9§ 11 ■ oj ..'-4
!Aof«l?
4 7 9 2 3
11. Whatisthevalueofg + g— j^+3^ + g — - +
- + -?
4^15*
12. If 8 J yards of broadcloth cost $29 f, how mudi will
10^ yards cost ?
13. If 8 J yards of broadcloth can be purchased for $29f)
how many yards can be purchased for $35^ ?
14. If 10^ yards of broadcloth cost $35^, how much will
8^ yards cost ?
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FRACTIONS. 21d
15. If $35^^ are paid for 10|^ yards of broadcloth, for how
many yards should $29 f be paid ?
16. If -}^ of a ton of hay costs $17.50, how much mU
two loads cost, one weighing f of a ton, and the other ^ of
/>< aton? -4
17. If 22 horses eat 41 1^ bushels of grain in 10^ days,
how many bushels will 21 horses eat in 9^ days ?
Solution. — As the answer is to be in bnshels of grain, we write 41^,
or -I J-&, as the number to be operated on. If 22 horses eat -^f^ bushels,
1 horse will eat j^ as much in the same time, which may be expressed
165
by making 22 a factor of the denominator ; thus, -— — » If 1 horse
will eat this quantity in 10^, or ^, days, in i day he will eat jV of
this, and in 2 half-days, or a day, he will eat twice this result, expressed
by making 21 a factor of the denominator, and 2 a factor of the numera-
tor ; thus, .* If 1 horse eats this quantity, 21 horses
will in the same time eat 21 times this, expressed by making 21 a factor
1 A^t \f O ^^ 01
of the numerator ; thus, -- — .t If 21 horses eat so much in
4 /\ 22 /\ ^1
1 day, in 9 J, or ^, days, they will eat ^ of this, which may be ex-
pressed by making 28 a factor of the numerator, and 3 a factor of tho
*t. 165 X 2 X 21 X 28^
denominator; thus, 4 X22X21X S "^
Hence the written work and answer are as follows : —
5
i$ t
bu. I|l><4^i|^><^ = 35 ba
4L X ^fi X fit X
Note. — It will be seen that in the solution, each condition of the
question was considered by itself, without reference to the other con*
ditions, and that the problem was dealt with as though composed of thf
following series of simpler problems :^-
1. If 22 horses eat 41 j- bushels of grain in a ^ven time, how msny
bushels would 1 horse eat in the same time ?
* This shows how many bushels 1 horse will eat in 1 day.
t This shows how many bushels 21 horses will eat in 1 day.
t This shows how many bushels 21 horses will eat in 9| dayi.
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216 FRACTIONS
2. If 1 hone eats the quantity obtained as the answer to the laJt
question in 10 j- days, how many bushels will he eat in 1 day ?
3. If 1 horse cats so much (the answer last obtained) in a day, how
many bushels will 21 horses eat in the same time f
4. If 21 horses eat so much (the answer last obtained) in 1 day, he w
many bushels will they eat in 9^ days ?
The student should notice that the question in each case has been^
" How many bushels ? " and that the denomination of each answer ham
been the same as that required for the final answer.
18. If 22 horses can be kept 10^ days on 41^ bushels of
grain, how many horses can be kept 9^ days on 35 bushels
of grain.
19. If 22 horses can be kept 10^ days on 41 J bushels of
grain, how many days can 21 horses be kept on 35 bushels
of grain?
20. If 21 horses eat 35 bushels of grain in 9 J days, how-
many hushels will 22 horses eat in 10^ days ?
21* If 21 horses can be kept 9^ days on 35 bushels of
grain, how many days can 22 horses be kept on 41^ bushels
of grain ?
22. If 21 horses can be kept 9^ days on 35 bushels of
grain, how many horses can be kept 10^ days on 41^ bushels
of grain?
23. If 7 men, by working 8 j- hours per day during 6 days
of the week, can earn $320^ in 5^ weeks, how many dollars
would 4 men earn in 4f weeks, by working 9 J hours per day,
and 5 days per week ?
24. K it costs $9.25 to dig a ditch 47 feet long, 1| deep,
and 3 feet wide, how much will it cost to dig a ditch 70^ feet
long, 2^ feet deep, and 4j^ feet wide ?
25. If, when flour is $6f per barrel, a six-cent loaf weighs
15 oz., how many ounces ought a ten-cent loaf to weigh when
flour is $9f per barrel ?
26. If a block of oak, 3f feet long, 1 foot wide, and f of a
foot thick, weighs 120f lb., how much will a block of pine, 8
^eet long, 2 feet wide, and 1§ feet thick weigh, pine being
only f as heavy as oak ?
27. If 9 men, by working 10 hours per day during 6 days
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FRACTIONS. 217
of tlie week, can in 4 weeks dig a trench 450 feet long, 3^
feet wide, and 2f feet deep, how many men, working 9^ hours
per day during 5 days of the week, can in 9 weeks dig a
trench 539 feet long, 6^ feet wide, and 2|^ feet deep ?
28. If 12 men, by working 9^ hours per day during 5
days of the week, can in 9 weeks dig a trench 539 feet long,
G^ feet wide, and 2|^ feet deep, how many weeks would it
take 9 men, working 10 hours per day during 6 days of the
week, to dig a trench 450 feet long, 3^ feet wide, and 2 J feet
deep ?
29. If 36 lb. of sugar are worth 24 lb. of coffee, and 22
lb. of coffee are worth 55 lb. of rice, how many pounds of
rice can be bought for 16 lb. of sugar ?
Preliminary Explanation. — As the answer is to be poands of rice,
we must write 55, the given number of pounds of rice, as the number
to be operated on, and must take care to express the values throughout
in pounds of rice. We should then have the following
Solution. — Since 22 lb. of coffee are worth 55 lb. of rice, 1 lb. of
coffee must be worth -^ of 55 lb. of rice, which may be expressed by
writing 22 as a denominator under the 55 ; and 24 lb. of coffee must be
worth 24 times the last result, which may be expressed by making 24 a
factor of the numerator. But as this is also the value of 36 lb. of sugar,
1 lb. of sugar must bo worth rf^ of this, which may be expressed by
making 36 a factor of the denominator; and 16 lb. of sugar must bo
worth 16 times the last result, which may be expressed by making 16 a
factor of the numerator.
The work would be written thus : —
5 ifi
lb. of rice ^^ ^ ^^ 21 ^^ - 261 lb. of rice = Ant.
^ 3
Note. — Examples like the above are usually solved by a process
called CoNJOiNBD Fbopobtion ; but as it is much less simple and con-
renient than the preceding, we omit it.
30. If 40 bushels of potatoes are worth 45 bushels of com,
and 18 bushels of com are worth 14 cwt. of hay, and 35 cwt.
of hay are worth 4 barrels of flour, how many barrels of
flour are 75 bushels of potatoes worth ?
19
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218 PRACTICE.
81. If 7 American yards are equal to 12 braces at Xieg»
horn, and 54 braces at Leghorn are equal to 45 braces at
Venice, how many braces at Venice are equal to 56 American
yards?
32. If John can earn as much in 7 days as William ran
earn in 9 days, and TVilliam can earn as much in 1 5 days as
Samuel can in 14, and Samuel can earn as much in 12 days
as Otb can in 8, and Otis can earn as much in 24 days as
Rofus can in 21, how many days will it take John to earn as
much as Rufus can earn in 49 days ?
33. Multiply 52 J by 7|, subtract 14f from the product,
and divide by {•
34. A man gave 4^ acres of land, worth $125.37 per acre,
in exchange for 75 barrels of flour. He sold f of the flour
at $9f per barrel, and the rest at $8^ per barreL How
much did he gain by the transaction ?
35. A garrison of 2000 men had bread enough to allow
each 24 oz. a day for 75 days ; but being besieged, it received
a reenforcement of 1600 men. How many ounces per day
can each man be allowed, in order that they may hold out 60
days?
36. Half of Arthur's money equals just ^ of William's,
and William has $187.47 more than Arthur. How much
money has each ?
15tS« Practice.
The following examples illustrate what is sometimes called
the Rule of Practice.
1. How much will 47 yd. 3 qr. 3 n. of broadcloth cost, at
$3,125 per yd. ?
Suffgestum, — Since 47 yd. 3 qr. 3 na. lacks bat 1 nail of being 48
fards, its cost mast eqa&l the cost of 48 yards, minus the cost of 1 nail.
WRITTEN WORK.
a,t=X) 3.125
48 X a =r b = ^150.000 =a cost of 48 yds.
TVofa = c = $ .195= " " Ina.
b — c = $149,805 =K " " 47 yds. 3 qr. 3 a*.
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PKACTICE. 219
2. How much will 26 A. 1 R. 25 sq. rd. of land cost, al
tl74.25 per acre ?
WHITTBN WORK.
a = $ 174.25 r= cost of 1 A.
26 X a = b = $4530.50 =3 «« « 26 A.
i of a = c = 43.562 = " " 1 R.
iofc=d= 21.781= " «208q.rd.
iofd=e=s 5.445= " " 5 sq. rd.
b + c+d + e = $4601.288 = " «« 26 A. 1 R. 25 sq. rd.
Note. — In dividing, in examples like the above, it is unnecesiMurf
to carry out tho work further than to mills.
3. How much will 13 lb. 4 oz. 10 dwt 8 gr. of silver cost
at $13.37 per lb. ?
4. How much will 9 T. 15 cwt 1 qr. 14 lb. of English
bar iron cost, at $83,625 per ton, reckoning the quarter at
28 lbs.? (See 34, c.)
5. How much will it cost to build a road 37 m. 5 fur. 80
rd. in length, at the rate of $2173.75 per mile?
WEITTEN WORK.
a = $ 2173.75 = cost of 1 m.
38 X a = b = $82602.50 = « « 38 m.
i of a = C = $ 543.437 = " »« 2 fur.
I of c = d = $ 67.929 = « " 10 rd.
b — c — d = $81991.134 = cost of 37 m, 5 fur. 30 rd.
6. How much will 22 hhd. 21 gal. 3 qt 1 pt. of wine cost,
at $47,875 per hhd., if each hogshead contains 63 gallons ?
7. How much will 24 bu. 1 pk. 6 qt. of wheat cost, at
$1,375 per bushel?
8. How much will 37 bu. 3 pk. 4 qt. of wheat cost at
$1,625 per bushel ?
9. How much will 14 cwt. 2 qr. 15 lb. of coffee cost, at
$12,625 per cwt.?
10. A man bought a farm, containing 137 A. 3 B. 30 sq.
rd., at $46.94 per acre ? How much did it cost him ?
11. A trader bought 184 bu. 2 pk. 6 qt. of Indian com, at
$.875 per bushel. How much did it cost him ?
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f20 PRACTICE.
12. How many rods will a man travel in 13 h. 20 m. 45
sec,, if he travels 1289 rods per hour?
13. What will 24 A. 1 R. 25 sq. rod cost, at $47.98 pet
acre?
14. What will 17 T. 15 cwt 2 qr. 7 lb. of railroad iron
cost, at $47.38 per ton ?
15. How much will 4 lb. 11 oz. 19 dwt 12 gr. of silver
cost, at $13.96 per pound ?
16. What will 428 bu. of oats cost, at $.49 per bushel ?
Suggestion. — At 49 cents per bashel, 428 bushels will cost 423 half
dollars, minus 428 cents.
17. How much will 43 volumes cost, at $1.99 per volume?
18. How much will 48 bbls. of flour cost, at $8.75 per
barrel?
Suggestion. — At $8.75 per barrel, 48 barrels will cost 48 times $9 —
46 times j- of a dollar.
19. How much will 32 yds. of cloth cost, at $1,875 per
yard?
Suggestion. — At $1,875 per yard, 32 yards will cost 32 times $2 —
82 times |- of a dollar.
20. How much will 747 lbs. of tea cost, at $.66| per lb. ?
21. How much will 8 cords of wood cost, at $4.94 per
cord?
22. How much will 4 acres of land cost, at $98.75 pei
acre?
23. What will 1728 yd. of cloth cost, at 1 s. 8 d. per yai-d?
Suggestion. 1 s. 8 d. = iV of ;£1.
24. What will 857 yd. of broadcloth cost, at 16 s. 8 d. per
yard?
Suggestion. 16 s. 8 d. = £1 -— 3 s. 4 d. = £1 — ^ of £1.
Or, 16 8. 8 d. = 10 s. + 6 s. 8 d. = ^ of £1 + i of £1.
25. What will 1478 reams of paper cost, at 13 s. 4 d. pei
ream?
^ggestion, 13 s 4 d. = £1 — 6 s. 8 d. « £i — |- of £1
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PRACTICE. 221
26. What will 737 yd. of black silk cost, at 17 s. 6 d. per
yard?
Suggestion. 2 8. 6 d. = ^ of £1 .
27. What will 138 yd. of silk velvet cost, at £1 5 s. 6 d.
£>or yard ?
WRITTEN WORK.
a = XI 38 = cost at jEI per yd.
i of a = b «= je 34 10 8. = " »» 5 8. per yd.
1.^ of b = c = X 3 9 8. = " " 6 d. per yd.
a -|- b + c = X175 19 s. == cost at £1 5 s. 6 d. per yard.
28. What will 13 coats cost, at £3 16 s. 8 d. each ?
WRITTEN WORK.
a = £13 s= cost at £1 each.
4 + a = b = £52 = « w £4 each.
{ of a = c = £ 8 13 8. 4 d. = cost at 3 s. 4 d. each.
b — c = £43 6 8. 8 d.
29. What will 24 tons of iron cost, at £8 2 s. 6 d. per ton?
30. What will 65 pieces of broadcloth cost, at £29 19 s.
8 d. per piece ?
31. What will 45 pieces of Irish linen cost, at £3 1 s; 4 d.
per piece ?
32. What will 96 tons of iron cost, at i£7 11 s. 4d. per
ton?
33. If 14 yd. 1 qr. 2 na. 1 in. of cloth cost £12 13 s. 8 d.,
what will 28 yd. 3 qr. 2 in. of cloth cost ?
Suggestion, 28 yd. 3 qr. 2 in. = twice 14 yd. 1 qr. 2 na. 1 in.
34. If 7 lb. 3 oz. 4 dr. of sugar cost 2 s. 9 d. 1 qr., what
will 21 lb. 9 oz. 12 dr. cost ?
35. If 17 bu. 1 pk. 2 qt. of com cost $9.39, how much will
51 bu. 3 pk. 6 qt cost ?
36. If 13 gal. 3 qt 1 pt. 1 gi. of molasses cost $3.87, how
much wiU 41 gal. 2 qt 1 pt 3 gi. cost ?
37. If 17 A. 3 R. 7 rd. of land cost $849.29, what will
88A. 8R.85rd. cost?
19*
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^22 RATIO AND PROPORTION.
88. If 98 yd. 3 qr. 2 na. of doth cost $44.88, what will
49 yd. 1 qr. 8 na. cost ?
39. 1£ 68 bu. 1 pk. G qt of wheat cost $114.48^ what will
22 bu. 3 pk. 2 qt cost ?
40. If 23 lb. 13 oz. 8 dr. of potash can be bought for
$2.12, how much can be bought for $16.96 ?
41. If 9 T. 14 cwt 2 qr. 17 lb. of hay can be bought for
$171.32, how much can be bought for $513.96 ?
SECTION XII.
RATIO AND PROPORTION.
NoTB. — If the teacher prefers it, the pnpil may omit this sectioa
and the two following it, till after he has mastered the sections on inter-
est and the subjects pertaining to business life.
ISO* Definitions and lUustrations of Ratio.
(a.) Ratio is the part which one number is of another, or
the quotient of one number divided by another : —
Thus, the ratio of 5 to 6 is f , or 1-^, because 6 equals f of 5, eqnalf
\\ times 5 ; or because 6 -f- 5 = f = 1-i-.
The ratio of 3 to 12 is ^^, or 4, because 12 equals -^^ of 3 = 4 timet
8 J or because 12 -*- 3 = -^ = 4
Note. — Some writers consider that the ratio of 5 to 6, or of 3 to
12, is the part which 5 is of 6, or 3 is of 12, instead of the part which 6
is of 5, or 12 is of 3, as given above.
The difference is not practically of so much consequence as wonld at
first appear to be the case, for the term ratio is almost invariably used
in some such connection as the following : " The ratio of 4 to 6 equals
the ratio of 10 to 15," where, by the first interpretation, we have f = W,
or f , = § ; and by the second, ^ = -J-^, or |, = ^, both of which are
manifestly true. It should, then, be borne in mind, that while the first
interpretation is the one usually adopted, the second may be si^bstitat^d
^r it, in any case where the change may seem desirable.
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KATIO AND PROPORTION. 223
\o,) A ratio can alw.iys be established between abstract
numbers, but it can only exist between concrete numberg
when they are of the same denomination ; for one quantity
can never be a fractional part of another quantity of a differ-
ent kind.
ThuSj 8 apples is no part of 4 pears, and hence has no ratio to it.
(c.) i]y the above illustrations, it appears that every ratio
is a true fraction, and may be written and dealt with as such.
(d,) Ratios are, however, usually expressed by writing one
number after the other, and placing two dots between them,
thus : — r-
The ratio of 4 to 6 = 4 : 6, or f .
The ratio of 9 to 7 = 9 : 7, or J.
(e.) The numbers which form any ratio are called terms
of the ratio ; the first number, or terra, is called the antece*
dent of the ratio ; and the second number, or term, is called
the consequent of the ratio.
Thus, in the ratio 9 : 7, 9 is the consequent, and 7 the antecedent
(/.) Ratios, like fractions, may be simple, complex, or
COMPOUND.
(g.) A SIMPLE RATIO IS the ratio of two entire numbers ;
as, 5 : 7, 4 : 9, or ^, |.
{h\ A COMPLEX RATIO is the ratio of two fractional
numbers ; as, | to 2^, 4f : 3-J, or -=., _1.
i ^
(t.) A COMPOUND RATIO is tjic indicated product of two
or more ratios ; as (5 : 7) X (8 : 3), or ^ of f , or |^ X §•
(y.) A compound ratio is usually expressed by writing the
mtios which compose it under each other : —
4:71
Thus, q . Q f expresses that the product of the two ratios, 4 : 7 and
9 : 8, is to be obtained ; or which is the same thing, it means J of |-,
or i X i
1^7. Beductio7i of Batios.
(a.) Since ratios arc really fractions, the principles in-
TolvAd in all operations apon them are precisely the same as
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224
BATIO AND PROPORTION.
those involved in tlie corresponding operations on other fimo
tions.
(b.) Hence, multiplTing or dividing both terms of a zatio
Dy the same number will not alter its value.
1. Reduce 4 : 6 to its lowest terms.
Thii
Solution. — Dividing both terms hy 2 gives 4 : 6
may be proved thus : 4 :6 = fs=J==2:8.
2:3.
Reduce each of the following ratios to its lowest ternos : —
2. 6 :
8. 12
4. 21
: 15.
: 35.
10.
11.
12.
13.
5.
6.
7.
24
16
8 :
: 48.
: 12.
6.
8.
9.
18
36
15.
54.
16 X 9 : 8 X 12.
15 X 3 X 7 : 35 X 9 X 4.
14 X 8 X 5 : 49 X 10 X 12.
20 X 7 X 45 : 9 X 35 X 10.
14. Reduce 4^ : 6^^ to a simple ratio.
First Solution. — Multiplying both terms hj 10, the least common
multiple of 5 and 10, (see 149, e.) gives 42 : 63 = 2 : 8.
Second Solution. 21 63
S 5 10
X 03( = 2 : 3. (See 149, c.)
Reduce the following complex ratios to simple ones : —
24.
15. 3^ : 4J.
16. .03 : 5.7.
17. |:f
18. ^ : 6f
19. .02 : 005.
20. 3f : 2f
21. 1| : 1^.
22. 2.0007 : 2000.7
23. .0025 : 2500.
5 : 8^
. Reduce 12 : 14 '
21 : 25J
>• to a simple ratio.
WRITTEN
3 3 2
fi X ifi X it : $X
r WORK.
2 5
Uxji$ = 9: 20.
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RATIO AND PROPORTION.
225
Pt«>f.
8 ,14 .25
2 2 5
$ X
9 : 20.
ii X M
3 S
20
9
Kcduce each of tLe following compound ratios to simple
28.
29.
80.
Note. — From the above, it is obvions that a compound ratio may
be reduced to a simple one, by making all its antecedents factors of the
new antecedent, and all its consequents factors of the new consequent,
and then cancelling all the factors common to both terms.
Itl8. Definitions and llltistrations of Proportion,
(a.) A PI? o PORTION is an equality of ratios.
(h,) Proportions may be simple, complex, or com-
pound.
(c.) A SIMPLE proportion cxpresses the equality of two
simple ratios.
{d,) A complex proportion expresses the equality of
two complex ratios, or of a complex and simple ratio.
(e,) A COMPOUND PROPORTION expresses the equality of
two compound ratios, or of a compound and simple ratio.
(/.) A proportion is expressed by writing two ratios one
after the other, and placing four dots between them.
Thus, 4 : 6 : : 12 : 18 is a proportion which expresses that the lano
of 4 to 6 equals the ratio of 12 to 18. It would be read, 4 ts to 6 as 13
M to 18, or 6 is the same part of 4 that 18 is of 12,
SJ 15^:: 2tV : ^i expresses that the ratio of 3^ to 5^ equals tha
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22t> RATIO AND rnOPOBTIOK.
ratio of 2^^^ to n-^. It wonid be read, 3^ t« to bi a$ 2iV f fo 3^, cr 5|
if t/te same part of 3 J that 3^ h of 2tV-
The compound proportion 4 : 7 ^
9 : 16 > : : 3 : 2, expresses that the rati«
14 : s)
of 4 X 9 X 14 to 7 X 16 X 3 equals the ratio of 3 : 2. It wonId be
read, 4 times 9 times 14 f& to 7 times 16 times 3 as 3 is to 2, or 7 times 16
ttMtes 3 is the same part of 4 times 9 times 14 <Aa< 2 t« o/* 3.
(i/.) The sign of equality is sometimes used instead of the
four dots.
Thus, instead of 4 : 6 : : 2 : 3, we may have 4:6 = 2:3.
(A.) Every proportion may be expressed as the equality
of two fractions.
Thus, we may express the first of the above by } = •{•§ , the second
by ^ == ?i-, and the third by J of ^^ of t\ = f .
({,) The outside terms (i. e., the first and fourth) of a pro-
portion are called the extremes, and the inside terms (i. e.,
the second and third) the means of the proportion.
Thus, in the first proportion above given, 4 i^nd 18 are the extremes.
6 and 12 are the means ; in the second, 3^ and 3-J- are the extremes
5^ and 2tV are the means ; in the third, 4 X 9 X 14 and 2 are th9
extremes, and 7 X 16 X 3 and 3 are the means.
1«S9« Method of finding a missing Term.
(a.) If any term of a proportion is omitted, it may easily
be supplied ; for from the nature of a proportion, it follows
that, —
First. Hie missing antecedent of any proportion must he
the same part of its consequent that the given antecedent is of
its consequent.
Second. The missing consequent of any proportion must he
the same part of its antecedent that the given consequent t$ of
its antecedent.
1. Find the missing term of the proportion 5 : 7 : : 25 : —
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RATIO AND PROPORTION.
227
Solution, — The missing term is the same part of 25 that 7 is of 5 ;
i. e., it is I of 25, which is 85. Hence, 5 : 7 : : 25 : l2^— = 85.
5
2. Find the missing term of the proportion 9 : 14 : : —
: 49
Solution,-^ The missing term is the same part of 49 that 9 is of U ;
o 9 X 49
i. 6., it is -^4 of 49, which is 31^. Hence, 9 : 14 : : --^ — = 31^ : 49.
8. Find the missing term of the proportion 4J 3^ : :
7i : -.
Solution, -» The missing term is the same part of 7-^ that 8 j> is of
4 J ; i. e., it is —
?i of 74. - "^ X ^ X i^^
= 5|.
Hence. 4fs3i::7i:?iA^=5f.
4. Find the missing term of the proportion
7 : 8 }■ : : 25 : — .
15 : I4J
Solution. — The missing term is the same part of 25 that 9 X 8 X
U is of 4 X 7 X IS ; L e., it ia
8 2 2 5
^ ^ ^ ^ ^Sf 25 =1><A44^^ =60.
4 X 7 X 15
HencO|
4 : 91
7: Sf
15 : 14 J
4x7t X i$
$
yX 8X14X25^
^ 4 X 7 X 15
Find the missing term of each of the following propo]>
tlons : —
5. 9 : 12 : : 6 : — .
6. ii : 10^ : : 26 : — .
7. 86 : 48 : : — : 60.
8. 62 : 42 : : _ : 3f
9. 4 : — :
10. 9f : —
11. — : 6 :
12. — : .9 :
20 : 28.
i 7 : 28.
18 : 52.
4.9 : .063
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223 RATIO AND PROPORTION.
r21 : 17^
15. ^ 42 : 32 > :: —
Ul : 54J
r. 03:. 007^1
16. < 3.5 : 500 V : : —
I .9 : 2.7J
64
SO.
160. Relations of the Terms.
(a.) From the foregoing explanations and illustrationsy we
K)k ly infer that, —
First Either extreme is eqtuzl to the quotient obtained By
dttiding the prodtict of the means by the other extreme.
Second. Either mean is equal to the quotient obtained by
dividing the product of the extremes by the other mean,
Ih,) Hence, in a proportion —
The product of the means is equal to the product of the
extremes,
101 • Practical Problems,
(a.) The forming of a proportion from the conditions of a
problem is called stating it.
(b,) In stating a proportion, it is customary to write the
number which is of the same denomination as the answer for
the third term.
1. If 8 books cost $4.32, what will 11 books cost?
SottUitm. — Since the answer is to be in dollars, we write $4.32 a«
the third term, and this being the price of 8 books, 11 books will cost
the same part of this that 11 is of 8, and we therefore mako 11 the
second ttsrm, nnd 8 the first Hence, the following statement . —
8:11:: $4.32 : $l:?i2Lii = $5.94 = Ans,
8
Note. — The above solution is really equivalent to the foUowing;
Since the answer is to be in dollars, we write $4.32, as the number on
which the operation is to be performed. If 8 books cost so much, II
books will cost V* o^ ^bis, which may be found by making 11 a fitetor
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RATIO AND PROPORTION. 229
of the numerator, and 8 a factor of the denominator. Hence, the fol-
lowing written work : —
2. If it takes 18 men 7 days to perform a piece of work,
how many men will it take to perform it in 9 days ?
Solution, — Since the answer is to represent the numher of men, we
write 18 men as the third term. But since it will take 7 times &s many
mon to do it in 1 day as it will to do it in 7 days, and i as many to do it
in 9 days as it will to do it in 1 day, it follows that the answer is th«
same part of 18 that 7 is of 9. Hence,
7 X 18
9 : 7 : : 18 : men t= 14 men = Ans.
9
^ Note. — When the pupil is familiar with the full form, he may
abbreviate, thus : Since the answer is to represent a number of men,
we write 18 men as the third term. But as this is the number of men
which it takes to do it in 7 days, it will take the same part of this num-
ber to do it in 9 days that 7 is of 9. This gives the same proportion as
before.
d. If it takes a man 7j> days, of lOj- hours each, to earn
$26j-, how many days, of 9-^ hours each, will it take him to
ean? the same sum ?
Solution. — Since the answer is to be in days, we write 7^ for the
third term. If, when the days arc loj* hours long, it takes so many
days, when they are 9-J" hours long it will take the same part of this
number of days that 10^- is of 9^, which will give 9-J for the first term,
and 10^ for the second. Hence, 9^ : 10^:: 7^ : — .
(c.) After having written the third term, we can tell in
what order to arrange the two remaining numbers, by observ-
ing that when the conditions of the question are such as to
require an answer greater than the third term, the larger
number will be the second term, and the smaller the first ;
and that when they are such as to require an answer less
than the third term, the smaller number will be the second
term, and the larger the first.
Thus, in the first example, having written $4.32 as the third term,
we may observe that the cost of 11 books will be greater than the cost
of 8 books, and that the ratio is 8 : 11.
20
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220 RATIO AND PEOPORTIOJf.
In the second example, harinf^ written 18 for the third term, tte mtf
observe that it will take a less number of men to perform the work ia 9
days than to perform it in 7 dav'*, and that the ratio is 9 : 7.
In the third example, havin;:,' written 7^ days as the third term, w©
may observe that it will take more days to earn the money when the^
are 9-^ hours long, than when they are 10^ hours long, and that the
ratio is 9| : 10^.
4. If 24 cows can be bought for $486, for how much can
17 cows be bought ?
5. K 17 cows can be bought for $344.25, for how mnch
can 24 cows be bought ?
6. If 24 cows can be bought for $486, how many can be
bought for $344.25 ?
7. If 17 cows can be bought for $344.25, how manj can *
be bought for $486 ?
8. How much would it cost to transport 9 cwt 2 qrs. 13 lb.
of merchandise from Providence to Boston, if it costs $6.43
to transport 12 cwt. 3 qrs. 11 lb. the same distance?
9. How many men will it take to perform a piece of work
in 6f days, which it will take 42 men 12^ days to perform ?
10. How many days will it take 81 men to perform a
piece of work which 42 men can do in 12^ days ?
11. How many men will it take to do a piece of work ia
12^ days which 81 men can do in 6§ days ?
12. How many days will it take 42 men to do a piece of
work which 81 men can do in 6§ days ?
13. John walks 3f miles per hour, and William walks 3 J
miles per hour. How many hours will it take John to walk
as far as William can walk in 8 hours ?
14. If 9J yards of cloth can be bought for $44|, how
many yards can be bought for $33 J ?
103* Problems in Compound Proportion.
1. If 6 boxes of soap, each holding 9 pounds, cost $4.59
how much will 11 boxes, each holding 12 pounds, cost?
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RATIO AND PROPORTION. 231
Solution, — Observing that the ar.swer is to be in dollars, we writ«
$4.59 as the third term. As this is the cost of 6 boxes of soap, 11
boxes will cost the same part of this that 11 is of 6, which may be ex*
pressed by makin;; 11 the second term, and 6 the first. But if boxes
each holding 9 pounds cost so much, boxes each holding 12 pounds
\vill cost the same part of this that 12 is of 9, which may be expressed
by making 12 a factor of the second term, and 9 a factor of the first.
Hemcc the following compound proportion : —
51 2
6 : 111
9 : 12J
4.59 : ^>g^ X 11 X Xi ^ ^^^^^
2. If 6 boxes of soap, each holding 9 pounds, can be
bought for $4.59, how mary boxes, each holding 12 pounds,
can be bought for $11.22.
Solution. — We first write 6 boxes for the third term. But as $1 1 .22,
or 1122 cents, will buy the same part of what $4.59, or 459 cents, will
buy, that 1122 is of 459, we make 1122 the second term, and 459 the
first. But these boxes hold 9 pounds each, and of boxes holding 12
pounds each, only t^jt as many can bo bought, which may be expressed
by making 9 a factor of the second term, and 12 a factor of the Qrst.
Hence the proportion
11
^^
459 : 11221 X H^^ X -_ ,,
12 : 9 j • • " • 4$^X i^ ~
$i ^
(a.) It will be seen by the above, that, ailer writing the
third term, we consider what ratio an answer depending on
any two similar conditions of the question would bear to that
third term, and that, after writing this ratio, we consider what
ratio an answer depending upon any other two conditions
eimilar to each other would bear to this, and so on till all the
conditions are considered. Then, by solving the resulting
compound proportion, the answer may be easily obtained.
(h,) If in any case the pupil is in doubt how to arrange
the terms of a ratio he may determine it by the method in/U
cated in 161 • c.
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232 RATIO AND rROPORTION.
Examples like those last explained are really eqaiyalent to
two or more examples in simple proportion.
Thus, the first is eqaivalcnt to the two following : —
1. If 6 boxes of soap cost $4.59, how much will 11 boxes cost ?
2. If a certain number of boxes of soap, each containing^ 9 ponndi^
cost the answer to the last question, how much will the same number of
hoxcs, each containing 12 pounds, cost ? (See note on 215th page.)
8. If it costs $36 to transport 14 tons of merchandise 54
miles, how much will it cost to transport 18 tons 49 miles ?
4. If 14 tons of merchandise can be transported 54 miles
for $36, how many tons can be transported 49 miles for $42 ?
5. If 14 tons of merchandise can be transported 54 miles
for $36, how many miles can 18 tons be transported for $42 ?
6. If 18 tons of merchandise can be transported 49 miles
for $42, how many tons can be transported 54 miles for $36 ?
7. If 18 tons of merchandise can be transported 49 miles
for $42, how many miles can 14 tons be transported for $36 ?
8. If it costs $42 to transport 18 tons of merchandise 49
miles, how much will it cost to transport 14 tons 54 miles ?
9. If 8 men can eani $216 in 3 weeks, how many dollars
can 12 men earn in 2 weeks ?
10. If it takes 24 pounds of cotton to make 2 pieces of
sheeting, each containing 83 yards, 1^ yard wide, how many
pounds of cotton will it take to make 11 pieces of sheeting,
each containing 27 yards, 1 J yard wide ?
11. If it takes 45 men 8 weeks, working 5^ days per
week, and 10 hours per day, to build a road 3 J miles long
and 4 rods wide, how many weeks will it take 63 men, work-
ing 4^ days per week, and 11 hours per day, to build a road
12§ miles long and 3 rods wide ?
12. A company of 40 men agree to perform a piece of
work in 50 days, but after working 9 hours per day for 30
days, they find that they have done but half the work. How
many more men must they employ, that, by working 10 hourj
per day, they may finish the job according to agreement ?
23. If 9 men, by working 8 hours per day, can mow 3t
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DUODECIMAL FRACTIONS. 233
acres of grass in 2^ days, how niany acres of grass can 5
men mow in 3 J days, by working 7j- hours per day ?
14. How many bottles can be filled from 12 casks of wine,
each cask containing 126 gallons, if 1344 bottles can be filled
from 3 casks, each cask containing 84 gallons ?
15. If, when land is worth $46| per acre, a lot of land, 35
rods long and 24 rods wide, is given for 8 piles of wood, each
45 feet long, 5 feet high, and 4 feet wide, how much ought
land to be worth an acre, when 3 lots, each 32 rods long, and
25 rods wide, are given for 21 piles of wood, each 37^ feet
long, 4J feet high, and 4 feet wide ?
SECTION XIII.
163. DUODECIMAL FRACTIONS.
(a.) A DUODECIMAL FRACTION, Or simply a DUODECIMAL,
18 a fraction whose denominator is 12, or some power of 12.
(b,) The denominator of a duodecimal is not written, but
is indicated by one or more marks, or accents, placed at the
right of the numerator, and a little above it
Thus, 4' = ^; 7'' = Ti¥; 11"' = tI^; &c
(c.) In reading duodecimals, twelfths are usually called
7RIMES ; one-hundred-^nd-forty-fourths are called seconds ;
&;c.
Thus, 4' is rcadj^bixr primes; 7'' is read, seven seconds; \V is read
deven thirds ; &c.
{d,) Duodecimals are employed in measuring lengths, sur-
faces, and solids ; so that the unit of measure is a foot of
either long, square, or cubic measure, according to the nature
of the thing measured.
(c.) Since a linear inch equals fV of a linear foot, a square
inch r^ of a square foot, and a cubic inch xtVf ^^ * cubic
20*
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£;>4 DUODECIMAL FRACTIONS.
foot, it follows that in long measure the inch is represented by
the prime, in square measure by the second, and in cubic
measure by the third.
(/) Duodecimals may be added, subtracted, multiplied,
and divided as other fractions are. In performing thes«^
operations, it is necessary to notice that a unit of any denomi-
nation equals 12 units of the next lower, and ^ of & unit of
the next higher denomination ; also, that 1' X 1' = iV X
iV = Tix. or 1''; that 1 ' X 1' = tIt X tV = ttW^ <>'
1'"; &c
163. Problems.
1. What ai-e the contents of a board 13 ft. 4' long, and 2
ft. 5' wide ?
Reasoning Process.— If the board was 13 ft. 4' long and 1 ft. wide, it
would contain 13 sq. ft. 4' ; but beinpr 2 ft. 5' wide, it roust contain 2^^
tioies 13 sq. ft. 4'. This gi^es the following written work : —
ft.
13 4
2 5
5 6 8''
26 8
32 2' S' = 32 sq. ft. 32 sq. in.
Explanation. — First, multiplying by .5' (i. e., by -^) we have 5'
times 4' = 20" = 1' 8'', (i. e., -^ times -^ = fW =?= iV + tIt-)
Writing the 8", wc have 5' times 13 ft. = 65', and 1' added, = 66' =
5 ft. 6'', (i. e., tV times 13 ft. = f §- ft., &c.,) which we write. Then
multiplying by 2, wc have 2 times 4' = 8' ; 2 times 13 ft. = 26 ft. Add-
ing these products gives 32 ft. 2' 8" for a result, which, being in squar«
measure, = 32 sq. ft. 32 sq. in.
Note. — The student should observe that in performing the work,
the multiplier is always to be rcp:ardcd as an abstract number. (See
74, g.) The expression, "multiplying the length by the breadth."
means simply that the number representing the length is to be mulli>
plied by the number representing the breadth.
2. What are the contents of a board 14ft. 5' long and 1 ft.
11' wide?
3. What are the contents of a blackboard 17 ft, 9' long
and 5 ft. 3 wide ?
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DUODECIMAL FRACTIONS. 285
4. What are the contents of a pile of wood 47 ft. 6' long,
8 ft. 9' wide, and 5 ft, 4' high ?
NoTB. — These examples can very frequently be performed more
easily by reducing the duodecimals to vulgar fractions. Thus, in the
last example, 47 ft. 6' = 47^ ft., 3 ft. 9 = sf ft., and 5 ft. 4' = 5^ ft.,
which, in conformity with 41^ Note, would give —
6 2
471 X 3f X 5i = ^|^^^3^ = 950cvi.&.= An,.
5. How many cubic feet of earth would be removed in
digging a cellar 15 ft. 9' long, 13 ft. 4' wide, and 9 ft. 8' deep?
6. How many square yards of carpeting will it take to
cover a floor 16ft. 6' long and 15ft. 4' wide ?
7. How many square feet and inches in the four walls of
a room 23 ft. 5' long, 18 ft. 9' wide, and 14 ft. 3 in. high ?
8. How many cords and cord feet of wood in a load 8 ft;,
long, 3 ft. 11' wide, and 5 ft. 7' high ?
9. How many yards of plastering are there in the top and
walls of a room 16 ft. 4' long, 13 ft. 2' wide, and 11 ft. 6'
high, allowing for three windows, each 5 ft. 8' high and 3 ft.
3' wide ; two doors, each 6 ft. 10' high and 2 ft. 9' wide, and
for a mop board, 9 inches wide, around the room ?
10. Mr. Jackson owned a garden, 137 ft. long and 112 ft.
6' wide, around which he laid out a gravel walk 4 ft. 8' wide.
How many square feet did the walk contain ?
Note. — The space used for the walk is supposed to be taken from
the contents of tlie garden.
Suggestion. — To understand this problem,
draw a figure like the one annexed. Then sup-
pose the walks along the sides, A B and D C,
to be first built. Each of them will be 137 ft.
long and 4 ft. 8' wide, and together they will be
equivalent to a walk 2 X 137, or 274 ft. long
and 4 ft. 8' wide. But in building these walks,
it is obvious that we have shortened each walk
to be built across the ends by just twice tJie width of the walk, i. e., by
twice 4 ^t. 8', which is 9 ft. 4'. This taken from 112 ft. 6 in., the width
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286 DUODECIMAL FRACTIONS.
of the garden, leaves 103 ft. 2', which is the length of the walk whicli
remains to be built across each of the two ends. Twice 103 ft. 2' =a
206 ft. 4', which, added to 274 ft., = 480 ft. 4', = the whole length of
the walk around the garden.
11. A man bought a garden spot, 143 fl. 4' long and 124
ft. 8' wide, and after leaving a space for a flower bed, 2 fl. 6'
wide, all around it, he laid out a walk, 3 ft. W wide, within
the flower bed, and extending around the garden. How many
feet did the walk contain ?
12. A man bought a lot of land, 97 ft. A' long and 73 ft
8' wide, at the rate of 6 cents per sq. ft. He built a tight
board fence, 6 ft. 2' high, around it, for which he paid 3 cents
per sq. ft How much did the land and fence cost him ?
13. How many yards of carpeting, 1^ yd. wide, will it
take to cover a floor 19 ft. 9' long and 15 ft. 6' wide ?
14. How many bricks, each 8 in. long, 4 in. wide, and 2
m. thick, will it take to build a wall 47 ft. long, 6 ft. 6' high,
and 1 ft. 8' thick ?
Suggestion. — Since 8 in. = § ft., 4 in. = ^ ft., and 2 in. = ^ ft.,
each brick contains § X i X i = sV solid feet, and it will take 27
bncks for each solid foot in the wall.
15. How many bricks, each 8 in. long, 4 in. wide, and 2 in.
thick, will it take to build the four walls of a house, 30 ft. 6'
long and 24 ft. 8' wide, the walls to be 1 ft. thick and 22 ft.
4' high, allowing for 1 door, 8 ft. high and 3 ft. 10' wide ; for
2 doors, each 8 ft. high and 3 ft. 6' wide ; for 12 windows,
each 6 ft. high and 3 ft. 8' wide ; and for 16 windows, each
5 ft. 8' high and 3 ft. 6' wide ?
Note. — The student should be careful to notice how the thickness
of the wall will affect his calculations. He must apply a principle
similar to that applied in the note to the lOth example.
16. A prison wall is built of Quincy granite, and extends
completely round the prison yard, except that a space is left
in one of its sides for an iron gate 12 ft. wide and 12 ft. high.
The outside length of the wall is 54 ft. 4', its breadth 49 ft.
6', and its height 18 ft. Its sides are 4 ft. 3' thick, and ita
ends 8 ft. 9' How many cubic feet in the wall ?
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ALLIGATION. 287
SECTION XIV.
ALLIGATION.
104 • Definitions and Eocplanations.
{a.) Merchants and others often find it convenient to mix
articles of different kinds together, so as to obtain a compound
differing in value from any of its ingredients. The various
problems connected with the subject are called Fboblems in
Alligation.
(5.) Questions in alligation are usually divided into two
classes, viz.: First, Alligation Medial, in which the
quantities of the several ingredients and their prices are
given, and we are required to find the price of the mixture
per pound, per gallon, or per bushel.
Second, Alligation Alternate, in which the pnces of
the various ingredients are given, and we are required to find
what quantities of each must be taken to make a mixture
having a given value per pound, per bushel, or per gallon.
16S. Problems.
1. A trader mixed together 6 lb. of coffee worth 10 cents
per pound, 4 lb. worth 8 cents per lb., and 7 lb. worth ItJ
cents per lb. How much was the mixture worth per lb. ?
Solution. 6 lb. at 10 cents are woi*th 60 cents.
^ (C i( Q i( (I (C 02 ((
7 «» « 16 ** " ** 112 "
Hence, 17 lb. are worth $2.04, and 1 lb. is worth ^ ot
$2.04, which is 12 cents = Ans,
2. A silversmith melted together 9 oz. of silver, 14 carats
fine; 6 oz., 18 carats fine; 12 oz., 22 carats fine; and 23 oz.,
24 carats fine. What was the fineness of the mixture ?
3. A dry goods dealer sold 25 yd. of sheeting at 9 cents
per yd. ; 30 yd. of shirting at 10 cents per yd. ; 11 yd. of
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288 ALLIGATION.
delaine at 18 cents per yd. ; 9 yd. of giDgbam at 22 cents
per yd. ; and 25 yd. of linen at 40 cents per yd. What was
the average price of the whole per yard ?
4. A merchant has sugars at 6, 7, 10, and 13 cents pei
pound, of which he wishes to make a mixture such that, by
selling it for 9 cents per pound, he will neither gain nor lose-
How many pounds of each kind must he take ?
Solution, — It is obvious that by selling the mixture for 9 cents per
pound, he will gain 3 cents on each pound which he puts in of the first
kind, and 2 cents on each pound of the second kind ; that he will lose
1 cent on each pound of the third kind, and 4 cents on each pound of
the fourth ; and further, that to make the mixture worth just 9 cents
per pound, he must take such proportions of the several kinds as will
make his gains equal his losses. Moreover, ho may take as many
pounds as he chooses of the kinds which cost less than 9 cents, provided
he takes enough of the others to counterbalance the gain on them.
Suppose that he takes 8 Ih. of the first kind, and 1 1 lb. of the second.
Then, since on 1 lb. of the first he gains 3 cents, on 8 lb. he will gain 8
times 3 cents, or 24 cents ; and since on 1 lb. of the second he gains 2
cents, on 11 lb. he mil gain 11 times 2 cents, or 22 cents, which, added
to the 24 cents, gives 46 cents as the sum of his gains. He must, there-
fore, take enough of the other kinds to lose 46 cents. Suppose he takes
10 lb. of that at 10 cents. Then, since on 1 lb. he loses 1 cent, on 10
lb. he will lose 10 times 1 cent, or 10 cents, and he must take enough
of that at 13 cents to lose 36 cents. Since on 1 lb. he loses 4 cents, ho
must take as many pounds to lose 36 cents as there are times 4 in 36
which are 9 times, fiencc, he may take 8 lb. at 6 cents, 11 lb *t 7
cents, 10 lb. at 10 cents, and 9 lb. at 13 cents.
The following is a convenient form of writing the work : —
Mean
Pricee of
Gain or loss
Quantities taken.
Sum of gains
price, ingredients
\wr lb.
or losses.
6 . .
. +3
8 lb. g. 24 cts.'
7 . .
. +2
11 « g. 22 " ,
46 cts. gain
10 . .
. — 1
10 " I. 10 " '
.13 . .
. — 4
9 " 1. 36 "
46 cts. loss.
Pivof. 8 lb. at 6 cents are worth $ .48.
11 " " 7 " « " $ .77.
10 " " 10 « " " $1.00.
V " " 13 « " « $1 17.
Making 38 lb., worth .... $3.42.
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ALLIGATION. 239
But 38 lb., at 9 cents per lb., would bring just $3.42, which showi
that the merchant would get the same sum by selling the mixture at 9
cents per lb. that he would by selling the ingredients separately, at theur
respective prices.
(c.) There is no limit to the number of answers wliich
may be obtained to such questions as the above ; for however
many or few pounds of any kind we take, we may take
enough of other kinds to counterbalance the gain or loss. In
the solution, we may as well consider first the number of
pounds to be taken of the kinds which cost less than the
mean rate, as of those which cost more.
{d.) Annexed is a part of the written work of two other
solutions to the above example. Let the student complete
and explain it, and also prove the correctness of his results.
Quantities taken. Sum of gains Quantities taken. Bam of gains
and losses. and losses,
lb. g. $.391 29 lb. g. $ .87\
7 lb. g. $.14 [ ^-^3 ^^°- 83 lb. g. $1.66 r ^2.53 gam.
6 lb. 1. $.051 lb. 1.
12 lb. 1. $.48] ^-^^ ^^^^' 26 lb. 1. $1.04
(e.) If it should be necessary to make a mixture oontaln*
ing a given number of pounds, we should first get an answer
to the question, as though no limit had been specified, and
then find how many times as much should be taken to give
the required quantity.
Suppose, for instance, that the above question had read, " How many
pounds of each kind must he take to make a mixtune of 100 lb. worth
9 cents per lb.," we should have the following additional work : —
The first solution gives a mixture containing 38 lb. ; and since 100 lb.
= 2i^ times 38 lb., we must take 2-tf times as much of each of the
former ingredients as before, which would give 2lf times 8 lb., or 21iV
lb. of the first ; 2^^ times 11 lb., or 28if lb. of the second ; 2^^ times
10 lb., or 26^y lb. of the third ; and 2i| times 9 lb., or 23-ft lb. of the
fourth. The proof is the same as though these quantities had been
originally selected.
5. A trader has coffees at 8, 10, 13, and 15 cents per lb.
How many pounds of each may he take to make a mixture
worth 12 cents per lb. ?
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240 INTEREST.
6. A trader has molasses at 22, 25, 29, and 33 cents per
gallon. How many gallons of each kind may he take to
make a mixture worth 26 cents per gallon ?
7. A trader has oils at $.95, $1.20, $1.42, and $1.60 per
gallon, of which he wishes to make a mixture worth $1.25 *
per gallon. How many gallons of each kind may he take ?
8. A trader wishes to mix 50 lb. of sugar at 7 cents
per lb., and 30 lb. at 10 cents, with such quantities at 9 and
6 cents per lb. as will make a mixture worth 8 cents per lb.
How many pounds of each may he take ?
9. A trader wishes to mix 40 lb. of tea at 40 cents per lb.,
80 lb. at 24 cents, and 50 lb. at 45 cents, with enough at 30
cents to make a mixture worth 35 cents per lb. How many
pounds of the last must he take ?
10. I have salt at 33, 37, and 50 cents per bushel. How
many bushels of each kind may I take to make a mixture of
100 bushels worth 40 cents per bushel ?
11. A farmer has oats worth 42 cents, barley worth 64
cents, rye worth 87 cents, and wheat worth $1.38 per bushel.
How many bushels of each kind may he take to make a mix-
ture of 200 bushels worth 75 cents per bushel ?
SECTION XV.
INTEREST.
1 66 • Introductory.
When a person hires an article of property of another, it
IB evident that, at the expiration of the time for which he
hires it, he ought to return it, and pay for its use. Moreover,
the sum paid for the use of the borrowed article should be
proportioned both to its value and the length of time it is
kept
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INTEREST. 241
For instance, if I hire two bouses, one of which is worth twice as
mucn as the other, I ought to pay twice as much per year for the first
fts for the second. If the values of the houses are alike, and one is
kept one half as long as the other, only one half as much ought to be
paid for the first as for the second.
To the Teadier. — It will be well to illustrate the above important
principles by questions similar in character to the following : —
If one man hires a horse to go a certain distance, and another hiics
one to go twice as far, how many times as much ought the second to
pay for its use as the first pays 1 What would have been the answer to
the above question, provided the second man had gone 3 times as far as
the first 1 4 times as far ? 3 J times as far ? ^ as far ? ^ as far %
§ as far ? &c. If the horses are hired by the hour, and the first man
keeps his horse three times as many hours as the second keeps his, how
many times as much ought he to pay for the use of it 1 What would
have been the answer had he kept it 5 times as long as the second ? 8
times as long ? 6 times as long ? ^ as long ? ^ as long ? ^ as
longi &c
Similar questions should be asked with reference to other objects
hired, as houses, money, &c., till the principle is fully understood.
167* Definitions.
(a.) Money is very frequently hired, and the sum to be
paid for its use is determined in accordance with the above
principles. (See 177th page, Ex. 24, Note.)
(5.) Money thus paid for the use of money is called In-
TEBEST.
(c.) The money used is called the Principal.
(d,) The principal and interest added together form tho
Amount, or entire sum due at any given time.
(e.) The interest of any principal is usually reckoned at a
certain per cent, i. e., a certain number of one hundredths of
that principal, for each year it is on interest. This per cent
is called the Rate per cent, or simply the Rate.
Note. — The term per centy from the Latin per centum, originally
meant by the hundred; but as it is now used in arithmetic, it means ons
hundredthM. Thus 6 per cent means tSiT) or .06 ; 4 per cen^ means .
21
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212 INTEREST.
T^Ht or .04 ; i per <ent means x^, or i of tw» or s^Jxr; &c. Thfs
term may be applied to any thing else as well as money ; and hence
the definition (often given in the school room) " so many cents on lOO
cents " is not a good one, any more than would be, so many biisIteJa or*
100 huiJitUy or so many yards on 100 yards. It is the more objcctionablo
because scholars are sometimes led by it, and by being often called upon
to Qse the term per cent in connection with money, to snpposc that it
feai lome necessary connection with cents, or with United States money
168. Legal Rate.
(a.) In most countries, laws have been passed regulating
in some way or other the rates of interest
(h) Such laws are called Usury Laws.
(e,) They commonly embrace the following particulars : —
First They t:ji the rate which shall be paid when no
special rate has been agreed upon by the parties. This is
called the Legal Bate.
Second. They forbid persons to receive interest at more
than some given rate.
Third. They impose penalties for their violation.*
(d,) Any excess over the rates allowed by these laws is
called Usury.
(e,) Li most states of the Union, the legal rate is also the
highest rate allowed by law, even on special contracts. The
exceptions to this are named in the following statement of
the legal rates of interest in the several states.
(/.) Li New York, South Carolina, Grcorgia, Michigan,
and Wisconsin, the legal rate of interest is 7 per cent per
year.
* It may not be amiss to remark that the laws regulating the rate of
interest are yery often disregarded, while the penalties for thek yiolation
are seldom imposed. Yery few men continue long in business without
paying or rec^ving interest at more than the legal rate. Money, haTing,
Uke every thing else, a yariable yalue, will bring what it is worth at the
time it is sold or let, and it seems as impossible to regulate by law the
price which shall be paid for its use, as to fix by law that which shall be
paid for the use of any othei article of property.
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INTEREST. 243
{t/,) In Alabama and Texas, it is 8 per cent per year.
(h,) In Louisiana, it is 5 i>er cent.
(t.) In California, it is 10 per cent.
(L) In all the other states, the legal rate is 6 per cent
per year.
(/.) By special agreement of the parties, interest may be
charged at the rate of 8 per cent per annum in Florida, Mis-
fiissippi, and Louisiana ; at the rate of 10 per cent in Arkan«
saSy Uiinois, Michigan, Iowa, and Ohio ; at the rate of 12
per cent in Wisconsin and Texas ; and at the rate of 18 per
ceni in California.
(/w.) On debts in favor of the United States, interest is
computed at the rate of 6 per cenL
(w.) In each state, interest is reckoned at the legal rate of
that sUite, unless otherwise specified.
(o.) In the United States, it is customary, when reckoning
interest, to regard a year as 12 montlis, and a month as 30
days. But in England, tlie year is regarded as 365 days, and
tlie number of days in the months considered are reckoned as
in the calendar.
(p.) In England, the legal rate is 5 per cent.
169. Interest for 2 Months, 200 ^MbntJis, ^c, at 6 per
cent.
(a.) If the interest of any sum for one year is 6 per cent
of that sum, for ^ of 1 year, or 2 months, it must be J of 6
per cent, or 1 per cent of it^
(b.) Therefore, at 6 per cent per year, the interest for 2
months is 1 per cent, or .01 of the principal, and mjiy be
found by removing the decimal point of the written principal
two places towards the left.
Thus, the interest of $75 for 2 months is $.75 ; of $364.30 is $3,643,
Ac, &c
What is the interest of each of the following sums fur 9
months ?
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244 INTEREST.
1. $84 3. $827.41* 5. $838.75
2. $687 4. $637.86 6. $3.86
7. What is the amount of each of the above sums foi 3
months ?
(c.) If the interest of any sum for 2 months is .01 of thai
sum, its interest for .1 of 2 months, which is 6 days, must bo
.1 of .01, or .001 of it
(cL) Therefore, at 6 per cent per year, the interest for 6
days is .001 of the principal, and may be found by removin/»
the decimal point three f laces to the left.
Thus, the interest of $987 for 6 days is $.987 ; of $439 is $.439 ;
of $8763.72 is $8,764 ; t or, carrying the values no farther than cents,
the interest of $987 is $.99 ; of $439 is $.44 ; of $8763.72 is $8.76 ; &c.
What is the interest of each of the following sums for 6
days?
10. $67 I 12. $1473.87
11. $36.75 13. $142
8. $586
9. $930
14 What is the amount of each of the above sums for 6
days?
(e.) If the interest of any sum for 2 months is 1 per cent
of that sum, its interest for 100 times 2 months, or 200
months, must equal 100 times 1 per cent, or 100 per cent of
it, which is the sum itself.
(/.) Therefore, at 6 per cent per year, the interest for 200
months, or 16 years 8 months, must equal the principaL
Thus, the interest of $47 for 200 months is $47 ; of $37.84 is $37.84 ;
of $23 is $23 ; &c.
What is the interest of each of the following sums for 16
yr. 8 mo., or 200 mo. ?
15. $38.73
16. $57
17. $67.95
18. $2763
19. $.28
20. $1.07
* The denominatious helow mills need not be given in the answer. In
deed, those below cents need not be given, if, when there are more than I
mills, 1 be added to the number of cents.
t Since 5.0fM)72 = more than i of a mill.
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INTEREST. 245
21. What is the amount of each of the above sums for 200
months ?
(y.) If the mterest of any sum for 200 months is equal to
that sum, its interest for -j^ of 200 months, or 20 months,
must equal -^ of it.
(h,) Therefore, at 6 per cent per year, the interest for 20
months, or 1 year and 8 months, is ^ of the principal, and
may be found by removing the decimal point one place to the
left
Thus, the interest of $63 for 20 months, or 1 yr. 8 mo., is $6.30 ;
of $78.60 is $7.86 ; of $8.79 is $.879, or, carrying the result only to
cents, $.88
What is the interest of each of the following sums for 20
months, or 1 yr. 8 mo. ?
22. $73.86 I 24. $673.70 I 26. $5.87
23. $578 I 25. $48.63 I 27. $63
28. What is the amount of each of the above sums for^O
months ?
170* Recapitulation and Inferences,
(a.) The importance of the above deductions is such Jis to
demand that they should be made perfectly familiar by all
who would become expert in casting interest at 6 per cent.
We therefore repeat them.
When interest is 6 per cent per year, —
First. The interest of any sum for 200 months^ or 16 yr. 8
mo.j will equal that sum.
Second. The interest of any sum for 20 months^ or 1 yr. 8
mo., wUl equal ^ of that sum, or as many dimes as there are
dollars in the principal.
Third. The interest of any sum for 2 months will equal
.01 of that ^um, or as many cents as there are dollars tV* the
principal.
Fourth. The interest (r any sum for 6 days will equal .001
of that sum, or as many mills as there are dollars in the prin*
eipeU,
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246 INTEREST.
(5.) It is therefore evident that for J of 200 months, the
interest will equal J of the principal ; for i of 200 months^ -I
of the principal ; &c
(c.) For ^ of 20 months, the interest will equal ^ of -j^,
or 5^ of the principal ; for ^ of 20 mo., ^ of -^y or -^jy of the
principal ; for 3 times 20 mo., 3 times .1, or .3 of the princi-
pal; &c.
((/.) For ^ of 2 months, the interest will equal i of ,01, or
y^ of the principal ; for J of 2 months, ^ of .01, or ^-^ of
the principal ; for 7 times 2 months, 7 times .01, or .07 of
the principal ; &c.
(c.) For J of 6 days, the interest will be ^ of .001, or
g^ViT of the principal ; for 3 times 6 days, 3 times .001, or
.003 of the principal ; &c.
171* Table showing Interest for convenient Fractional
Parts of 200 months, 20 months, S^e,
At 6 per cent per year, the interest for —
200 mo., or 16 yr. 8 mo., = prin.
i of 200 mo., or 100 mo., or 8 yr. 4 mo., = i of prin.
■J of 200 mo., or 66§ mo., or 5 yr. 6 mo. 20 da., =s -J of prin-
i of 200 mo., or 50 mo., or 4 yr. 2 mo., = i of prin.
i of 200 mo., or 40 mo., or 3 yr. 4 mo., = i of prin.
i of 200 mo., or 33^ mo., or 2 yr. 9 mo., 10 da., == i of prin.
•J of 200 mo., or 25 mo., or 2 yr. 1 mo., = J of prin.
iV of 200 mo., or 20 mo., or 1 yr. 8 mo., = xV of prin.
iV o{ 200 mo., or 16f mo., or 1 yr. 4 mo. 20 da.. = iV of P"n-
iV of 200 mo., or 13^ mo., or 1 yr. 1 mo. 10 da. = ^ of prin.
iV of 200 mo., or 12 J mo., or 1 yr. 15 da., = iV of prin.
i of 20 mo., or 10 mo., s= i of iV of prin.
i of 20 mo., or 6§ mo , or 6 mo. 20 da., =x "J of i\r of prin.
J of 20 mo., or 5 mo., = i of -jV of piio*
I of 20 mo., or 4 mo., = i of iV of prin.
i of 20 mo., or 3^ mo., or 3 mo. 10 da., = J of i^ of prin.
^ of 20 rao., or 2^ mo., or 2 mo. 15 da., = ij- of -jV of prifl.
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INTEllEST.
l\r of 20 mo^ or 2 mo., =
lV of 20 mo., or if mo., or 1 mo. 20 da., =
f*5- of 20 mo., or 1^ mo., or 1 mo. 10 da., =
^ of 2 mo., or 1 mo., =
J of 2 mo., or § of 1 mo., or 20 da., =
I of 2 mo., or J of 1 mo., or 15 da., =3
J of 2 mo., or f of 1 mo., or 12 da., =
^ of 2 mo., or -i" of 1 mo., or 10 da., =
iV of 2 mo., or "5- of 1 mo., or 6 da., =s
1^ of 2 mo., or 1^ of 1 mo., or 4 da., =
J of 6 da., or 3 da., =
■J- of 6 da., or 2 da., =
J of 6 da., or 1 da., =
247
"riy of 1^ of prin.
tV of iV of prin.
tV of tV of prin.
i of T^ of prin.
i of lijj of prin.
i of Thj of prin.
i of lixT of prin.
i of T^Tj of prin.
tV of Tihr of prin
tV of ihr of prin.
i of iTjVir of prin.
i of txjW of prin.
i of luW of prin.
17S. Questions on the preceding Table.
1. How long musi
t a principal be on interest that the inter.
est may equal ^ of it ?
2.
iV?
15.
V
27.
iV?
3.
T^^?
16.
iof^?
28.
t'^?
4.
iof^?
17.
iof tV?
29.
*of A?
5.
iof^?
18.
tV?
30.
iof^V?
6.
^?
19.
-rb?
31.
^V?
7.
iV?
20.
T^?
32.
iVofiV?
8.
iVofA?
21.
?iT7?
33,
rirr?
9.
.01?
22.
i of .01 ?
34
J of .01 ?
10.
i of .01 ?
23.
TsW?
35.
Taiy •
11.
aiT?
24.
Winr'
36.
nilT?
12.
.001 ?
25.
fttW'
37.
* of y^^^ ?
13.
"ffTjVl) ?
26.
W^TT?
38.
i of .001 ?
14.
i?
173* Applications of the foregoing Table.
1. What is the interest of $156.96 for each time mentioned
in the taUe?
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248 INTEREST.
Answer. — The interest of SI 56.96 for 200 mo. = $156.96 ; for lOi
mo., or 8 yr. 4 mo., = ^ of $156.96 j for 66§ mo., or 5 yr. 6 mo. 20 da^
■=i of $156.96; &c
The interest of $156.96 for 20 mo. = $15.696 ; for 10 mo. = J of
#15.696 ; for 6 1 rao., or 6 mo. 20 da., = ^ of $15,696 ; &c.
The interest of $156.96 for 2 mo. = $1.57 ; for 1 mo. =i of $1.57;
for 20 da. = i of $1.57 ; &c.
KoTE. — The object in giving the answer in the above form is to
direct the pupirs attention exclusively to the method of computing the
interest ; but as soon as he has had practice enough to enable him to
tell at once what part of the principal, or of some convenient part of
the principal, the interest for any of the above-mentioned times is, he
should compute the interest in each case. Thus, the interest of $156.96
for 100 mo., or 8 yr. 4 mo., = i of $156.96, which is $78.48 j for 66^
mo., or 5 yr. 6 mo. 20 da., = ^ of $156.96, which is $52.32 ; &c.
After a little practice, the final answer may be read at once.
Thus,—
The interest of $156.96 for 100 mo., or 8 yr. 4 mo., is $78.48 ; for
«6§ mo., or 5 yr. 6 mo. 20 da., is $52.32 ; &c.
What is the interest for each of the above-named times of,
2. $48? 4. $72? I 6. $24.96?
3. $42.24? 5. $144? 7. $35.42?
8. What is the interest of $36 for 3 yr. 4 mo. ?
9. What is the interest of $24.72 for 16^ mo. ?
10. What is the interest of $75 for 2 yr. 9 mo. 10 da. ?
11. What is the interest of $54 for 5 >t. 6 mo. 20 da. ?
12. Wliat is the interest of $324 for 33^ mo. ?
13. What is the interest of $231.12 for 1 yr. 15 da. ?
14. What is the interest of $42.24 for 2 yr. 1 mo.?
15. What is the interest of $500 for 66§ mo. ?
16. What is the interest of $42.24 for 1 yr. 8 mo. ?
17. What is the interest of $150 for 6 mo. 20 da ?
18. What is the interest of $4.80 for 5 mo. ?
19. What is the interest of $17.70 for 2 mo. 15 da. ?
20. What is the interest of $87.18 for 1 mo. 20 da. ?
21. What is the interest of $537 for 2 mo. ?
22. What is the interest of $288 for 10 da. ?
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INTEREST. 249
23. What is the interest of $96.34 for 20 da, ?
24. What is the interest of $675 for 2 da. ?
25. What is the interest of $438.74 for 3 yr. 4 mo. ?
2G. What k the interest of $73.87 for 1 yr. 15 da. ?
27. What is the interest of $144 for 1 yr. 1 mo. 10 da. ?
28. What is the mterest of $56 for 4 yr. 2 mo. ?
29 What is the interest of $1728 for 1 yr. 4 rao. 20 da. ?
30. What is the interest of $327.29 for 16 yr. 8 mo. ?
31. What is the mterest of $44.20 for 10 mo. ?
32 What is the interest of $57.60 for 3 mo. 10 da. ?
33. What is the interest of $43.65 for 4 mo. ?
34. What is the interest of $7.65 for 1 mo. 10 da. ?
35. What is the interest of $97.20 for 6 mo. 20 da. ?
36. What is the interest of $237.50 for 1 mo. ?
37. What is the interest of $541 for 5 da. ?
38. What is the interest of $9741 for 6 da. ?
89. What is the interest of 882.47 for 3 da. ?
40. What is the interest of $32.76 for 2 yr. 9 mo. 10 da. ?
41. What is the interest of $93.27 for 6 mo. 20 da. ?
174. Interest for various Times.
(a.) In computing interest for other times than those al-
ready mentioned, it is usually most convenient to divide the
time intb parts, as illustrated below.
(6.) The student should bear in mind that in the forms of written
work here, as elsewhere, the letter a, b, c, &c., are used merely to indi-
cate how the numbers standing opposite them have been obtained. la
oiactical work they should be omitted.
1. ^VTiat is the interest of $196.72 for 8 mo. 20 da. ?
First Solution,
a = $196.72 = Principal.
^ of a = b =s 6.557 = Int. for 6 mo. 20 da
.01 of a = c = 1 .967 = Int. for 2 mo.
b + c = $ S.524 = Int. for 8 mo. 20 da.
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15^ INTEREST.
Second Solution.
a = $196.72 = Principal
j04 of a =5 b =s 7.869 = Int for 8 mo.
ifijf of a =s c =3 .655 =3 Int. for 20 da.
b + c = I 8.524 = Int. for 8 mo. 20 da.
2"hird Solution.
a = $196.72 = Principal.
2^5 of a ae b = 9.836 == Int for 10 mo.
y^ of a =:s c =s 1.31 1 = Int. for 1 mo. 10 da.
b — c = $ 8.525 = Int. for 8 mo. 20 da.
Fourth Solution.
Since at 6 per cent the interest for 1 month equals i of 1 per cent
of the principal, the interest for 8 months must equal 4 per cent of the
principal ; and since the interest for 1 day equals ^ of .001 of the prin-
cipal, the interest for 20 days must equal ^ of .001, or .003^ of the
principal Hence, the interest of $169.72 for 8 mo. 20 da. = .04 + .OOSi
1= .043-i- of $169.72, which, found by the usual method, gives $8,525
Bs interest for 8 mo. 20 da.
2. Wliat is the amount of $649.37 for 17 mo. 15 da. ?
First Solution.
a = $649.37 = PrincipaL
tV of a = b = 54.114 = Int for 16 mo. 20 da.
jJqp * of b = c = 2.705 = Int. for 25 da.
a + b + c =3 $706,189 = Amt. 17 mo. 15 da.
Second Solution.
a s= $649.37 = Principal.
iV of a = b = 40.585 = Int. for 12 mo. 15 da.
iV of a = c = 16.234 = Int for 5 mo.
a + b + c = $706.1 89 == Amt for 17 mo. 15 da.
• Since 16 mo. 20 da. =: 500 da.
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INTEREST. 2S1
Third Solution,
a = $649.37 = Principal.
^ of a =3 b =: 64.937 = Int. for 20 mo.
i of b = c = 8.117 = Int for 2 mo. 15 da.
a + b — c = $706,190 = Amt. for 17 mo. 15 da.
Fourth Solution.
a = $649.37 = Principal.
sV of a = b = 32.468 = Int. for 10 mo.
1% of a, or ]jr of b, = c = 16.234 = Int. for 5 mo.
gV of ftj or ]J- of c, = d = 8.117 = Int. for 2 mo. 15 da
a + b + c + d= $706,189 = Amt. fof 17 mo. 15 da.
Fifth Solution.
Since at 6 per cent the interest for 1 month is ^ of 1 per cent of th«
principal, the interest for 17 months must equal -y- of 1 per cent=a
.08^ = .085 of the principal ; and since for 1 day the interest equals i
of .001 of the principal, for 16 days it must equal -^ of .001, or .002?
of the principal. Hence, the interest of $649.37 for 17 mo. 15 da. =»
.085 + .002f = .087f of $649.37, which, found by the usual method,
gives $56,819, the interest, which, added to the principal, gives the
amount, $706,189.
Note. — Many other solutions might have been given to the above
questions, but as they would all involve similar principles, it is unneces-
sary to add them. Every question in interest admits a great variety of
solutions, and the pupil should examine it carefully to determine which
he will adopt. Practice will enable him to select a good method at
once. One process may be applied to test the correctness of a result
obtained by some other. We may remark, that, as a general thing, it
is better to divide than to multiply, for in division we have tc consider
no denomination below the lowest we wish to have in the answer.
3. What is the interest of $857.63 for 3 mo. 16 da. ?
4 Wliat is the interest of $875.37 for 1 mo. 26 da. ?
5. What is the interest of $93.75 for 9 mo. 29 da. ?
6. What is the interest of $178.43 for 16 mo. 14 da. ?
7. What is the mterest of $343.65 for 13 mo. 16 da. ?
8. What is the interest of $237.64 for 19 mo. 24 da. ?
9. What is the interest of $478.96 for 17 mo. 26 da.
10. What is the interest of $375.81 for 22 mo. 15 da. ?
11. What is the interest of $58.27 for 96 mo. 20 da. ?
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252 INTEREST.
12. What is the interest of $5789 for 29 mo. 29 da. ?
13. What b the interest of $80.32 for 7 mo. 19 da. '
U. What is the interest of $175 for 1 mo. 17 da. ?
15. What is the interest of $326 for 8 mo. 23 da. ?
16. What is the interest of $27.96 for 1 yr. 3 mc 13 da. ?
17. What is the amount of $578.31 for 3 yr. 7 mo* 28 da. ?
18. Wliat is the amount of $724.16 for 7 yr. 2 mo. 11 da.?
19. What is the amount of $4369.87 for 3 mo. 26 da. ?
20. What is the amount of $25.50 for 9 mo. 27 da. ?
21. What is the amount of $117.58 for 3 yr. 1 mo. 18 da. ?
22. What is the amount of .$313.27 for 6 mo. 9 da. ?
23. What is the amount of $57.75 for 9 mo. 1 da. ?
24. What is the amount of $35.86 for 11 mo. 25 da. ?
25. What is the amount of $17.64 for 1 yr. 1 mo. 13 da. ?
26. What is the amount of $378.51 for 1 yr. 5 mo. 17 da. ?
175. ComptUaiion of Time, and Application to Problems*
(a.) In business transactions, it is usually necessary Xxi
compute the time during which money has been on interest ;
that is, the time between the dates on which interest began
and ended.
(^.) The usual method of doing this is to reckon the num-
ber of entire years, then the number of entire calendar
months remaining, and then the remaining days.
(c.) The year is (as before) regarded as 360 days, or 12
months of 30 days each, and each entire calendar month as a
month of 30 days ; but the days which are left after reckoning
the years and months are determined by counting them ac-
cording to the number in the months in which they occur.
1. What is the time from Jan. 17, 1845, to June 28, 1849?
Solution, — From Jan. 17, 1845, to Jan. 17, 1849, is 4 years; from
Jan. 17 to June 17 is 5 months; from June 17 to Jane 28 is 11 days.
Therefore the required time is 4 yr. 5 mo 11 da.
2. What is the time from April 27, 1846, to Feb. 18,
1851?
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tKTEREST. 253
Solution. — From April 27, 1846, to April 27, 1850, is 4 yeartJ*, from
April 27, 1850, to Jan. 27, 1851, is 9 months ; January having 31 dayi,
there are 4 days left in it, which, added to the 13 in February, give 17
days. Therefore the required time is 4 yr. 9 mo. 17 da.
3. What is the time from Sept. 24, 1849, to March 20,
1852?
Solution. — From Sept 24, 1849, to Sept. 24, 1851, is 9 /ears ; from
6ept. 24, 1851, to Feb. 24, 1852, is 5 months; 1852 being leap year,
February has 29 days ; hence, there are 5 days left in it, which, added
to the 20 in March, give 25 days. Therefore the required time is 2 yi.
5 mo. 25 da.
Note. — This method of computing the time, though the one usual-
ly adopted by business men when interest is computed for months and
days, is unequal in its operation ; for the calendar months, though vary-
ing in length from 28 to 31 days, are all reckoned as months of 30 days
each. Hence, the interest of a sum during the month of February will
be as much as during either March or April, though February contains
3 days le§s than March, and 2 less than April.
By this method, the interest on four notes dated respectively on the
28th, 29th, dOth, and 31st of any one month, and paid on any one day
between the 1st and 28th of March, of any year except leap year,
would be computed for the same time. Suppose, for instance, that they
are dated in October, 1850, and paid March 15, 1851. Then for the
first note dated Oct. 28, the time will be found without difficulty to be 4
mo. 15 da. In calculating the time on the others, we proceed thus:
Since there are not as many as 29 days in February, 1851, we reckon
from the 29th, 30th, or 31st of October to the last day of February as 4
months, to which adding the 15 days in March gives 4 months and 15
days as the time, in each case. The restriction with reference to notes
paid in leap year is necessary simply because February has then 29
days. The proposition will always be true of notes dated on the 29th,
SOth, and 31st, of any month, and paid at any time between the 1st and
28th of March.
Again. Four notes dated respectively on the 28th, 29th, 30th, and
8l8t of August of any year except the one immediately preceding leap
year, and payable in 6 months, would all become duo on the same day.
The only strictly accurate method of reckoning time is to actually
count the days in each month we consider. Thus, to reckon the tim^
from October 28, 1850, to March 15, 1851, we proceed as follows : Fronk
October 28th to 31st, is 3 days ; to which adding the 30 days in Noven*-
ber, the 31 in December, the 31 in January, the 28 in February, and tbe
15 in M^rch, gives 138 days as the true time between the two dates
22
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25-4 INTEREST.
In England the time is A.vmjB computed in this ^aj, as it is in tfa«
country when notes are payable at the end of a certain number of dayt-
4. What is the time fn.tn June 23, 1850, to June 3, 18.'^2 ?
Answer. 1 yr. 11 mo h da.
5. What is the time from May 13, 1847 to Oct 8, 1851 ?
6. What is the time from Jan. 31, 1851 .to March 23,
1852?
7. What is the time from Nov. 17, 1849, to Dec 12, 1851 ?
8. What is the time from Dec 31, 1848, to July 6, 1850 ?
What is the interest —
9. Of $787.36 from May 3, 1843, to Dec 17, 1845 ?
10. Of $54.76 from Feb. 14, 1840, to June 2, 1844?
11. Of $476.35 from June 30, 1847, to Dec 28, 1850?
12. Of $638.29 from May 31, 1851, to Oct. 7, 1852 ?
13. Of $4937.56 from Dec 19, 1843, to Feb. 16, 1847 ?
14. Of $481.74 from Jan. 29, 1847, to March 25, 1851 ?
15. Of $587.60 from Jan. 31, 1850, to July 18, 1852?
What is the amount —
16. Of $947.84 from May 15, 1850, to June 13, 1851 ?
17. Of $748.67 from Dec 14, 1849, to May 4, 1851 ?
18. Of $1546.61 from April 9, 1847, to June 1, 1851 ?
19. Of $917.68 from June 5, 1842, to Jan. 1, 1850?
20. Of $8396.58 from April 30, 1847, to March 22, 1850 ?
21. Of $1449.13 from Dec 31, 1850, to March 5, 1852 ?
22. Jan. 15, 1852, George W. Pratt borrowed $237.50 of
A. N. Johnson, and Feb. 13, 1852, he borrowed $438.75
more, agreeing to pay interest at 6 per cent per year. Ha
paid the debts March 8, 1852. What was their amount ?
23. I have three notes against Arthur Sumner, viz., one
for $548.17, dated Jan. 1, 1851, another for $679.18, dated
Jan. 27, 1852, and another for $376.89, dated May 31, 1852.
What amount will be due on all of them July 8, 1852 ?
24. Jan. 1, 1851, I borrowed $3468, with which I pur^
chased flour at $6 per barrel. I sold the flour March 17,
1851, for $6.50 per barrel, cash. Interest being reckoned at
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INTEREST.
255
6 per cent, did I gain or lose by the transaction, and how
much ?
170. Interest hy Days,
(a,) Many business men always reduce the time to days,
and conmute the interest by the method illustrated below. As
a generalihing, however, this method is not so convenient as
the preceding
{h.) Since, ^6 per cent per year, the interest for 1 day ;
I ^^ ttjVtt o^ theWincipal, it follows that the interest for/my
number of days mu^t be ^ as many thousandths of tWprin-
cipal as there are day^
(c.) We may, therefore, find the interest for j&y number
of days, by multiplying thesprincipal by ^ of jKe number of
days, and removing the poi^t three plaj)^ farther to the
left.
It will make no difference with the ran^ whether we multiply by i
of the number of days, or by the numl^igrV days and divide by 6 ; but
it will usually be better to diride before mulu|>lying.
1. What is the interest (^^437.62 for^4 mo. 3 da. ?
Solution. — Since 4 mo. 3 da. = 123 da., the reqiured interest most
be i of .123 of the princuml, whi/iSfi^ is .0202- of thXprincipal. Tht
work carried to mills woi^ be wr^ten thus : —
/ $43^62
2.
8.
4.
5.
6.
7.
8.
9.
What is the
What is the
What is the
What is the
What is the
What is the
What is the
What is the
75J
J2l8
$4^70
interest of
interest of
interest of
interest of
interest of
interest of
interest of
interest of
= Ans.
$54.57 for 4 mo. 20 da. ?
$397.42 for 8 mo. 15 da. ?
$231.48 for 7 mo. 12 da,?
$438.64 for 5 mo. 24 da. ?
$281.87 for 5 mo. 9 da. ?
$581.21 for 6 mo. 24 da. ?
$83.25 for 2 mo. 21 da. ?
$98.37 for 6 mo 18 da. ?
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256
INTEREST.
17T* Liter est hy Dollar Sy for Months and convenient p€srt§
of a Month,
(a.) TVe can frequently compute interest with great ease
bj first finding the interest for 1 day, 1 month, or 1 year, and
getting the required interest from this.
{b.) When the interest for any of the above times is any convenient
snm, as 1 dollar, 1 dime, 1 cent, 1 mill, i dollar, i dollar, &c., the pro-
posed course will bo particularly advantageous.
(c.) Since, at 6 per cent per year, the interest of any sum
for 1 month is 5^ of that sura, —
1. The mterest of $200 is $1 per month.
2. The interest of $20 is $.10, or 1 dime per month.
8. The interest of $2 is $.01 per month.
1. What is the interest of $200 for 7 mo. 15 da. ?
Solution. — The interest of S200 is $1 per month j therefore for 7 mow
15 da., or 7i mo., it must be 74 dollars, or $7.50.
What is the interest of $200 for —
2.
5 mo.?
7.
8 mo. 12 da. ?
3.
8 mo.?
8.
9 mo. 15 da. ?
4.
6 mo. 10 da. ?
9.
1 yr. 7 mo. ?
5.
1 yr., or 12 mo. ?
10.
8 yr. 8 da. ?
6.
2 yr. 1 mo., or 25 mo. ?
What is the interest for each of the above-mentioned
times —
11. Of $2? * I 12. Of $20? I 13. Of $2000?
14. What is the interest of $100 for 7 mo. ?
Solution, — Since the interest of $200 is 1 dollar per month, the in-
terest of $100 must be a half dollar per month, and 7 half-dollara, oc
$3.50, for 7 months.
What is the interest of $100 for —
15.
16.
17.
18.
9 mo.?
lyr.?
8 yr. 4 mo. ?
15 da. ?
19. 20 da.?
20. 8 mo. 15 da. ?
21. 2 yr. 9 mo. ?
22. 8 mo. 10 da. ?
23. lyr. 2mo. 20da.F
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INTEREST. 257
What is the interest for €ach of the above-mentioned
24.
Of $10?
27.
Of $50?
30.
Of $25 ?
25.
Of $1?
28.
Of $5?
31.
Of $250 ?
26.
Of $1000?
29.
Of $500 ?
82.
Of $2500 ?
178. Interest hi Dollars, when the Time is in Days, or
Months and Days.
(a.) Since the interest of any sum for 6 days is .001 of
that sum, the interest of 1 dollar for 6 days must be .001 of
1 dollar, which is 1 mill. If the interest of 1 dollar for 6
days is 1 mill, its interest for 1 day must be ^ of 1 milL
Therefore, the interest of $1 is ^ of 1 miU per day.
(b.) From this we have the following statements : —
At 6 per cent, —
1. The interest of $Q is 1 mill per day.
2. The interest o/" $60 is 1 cent per day.
3. The interest of $600 is 1 dime per day.
4. The interest of $6000 is 1 dollar per day.
Note. — When the interest is required for months and days, "we may
reduce the months to days, and then apply the above ; or we may find
the interest for the months, as in the last article, and then find it for the
days, as above. The pupil should 'remember that the interest of $6 is 3
cents per month, of $60 is 3 dimes per month, and of $600 is 3 dollars
per month.
What is the interest of $6 xor —
1. 15 da. ?
2. 12 da. ?
3. 20 da- ?
4. 1 mo. 3 da. ?
5. 3 mo. 10 da. ?
6. 7 mo. 15 da. ?
7. 2 yr. 3 mo. 5 da. ?
8. 1 yr. 4mo. 11 da.?
9. 4 yr. 7 mo. 20 da. ?
What is the interest for each of the above-mentioned
times —
10. Of $60? 1 11. Of $600? 1 12. Of $6000?
13. What is the interest per day of $750 ?
22*
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258
INTEREST.
SolvtioH. — The interest of $6000 being 1 dcMar per day, the intenMf
of $750, which is i of $6000, mast be ^ of 1 dollar per day.
What is the interest per day —
, 14. Of $1000? 18. Of $10? 22. Of $30?
15. Of $2000? 19. Of $150? 23. Of $180 ?
16. Of $4000? 20. Of $15? 24. Of $800?
17. Of $1200? 21. Of $120? 25. Of $500?
26. What is the interest per month of each of the above
sums?
What is the interest of $3000 for —
27.
28.
29.
30.
31.
8 da.?
15 da.?
1 mo. 10 da. ?
20 da.?
2 mo. 10 da. ?
82.
33.
84.
35.
27 da. ?
9 da.?
25 da.?
11 da.?
86. What is the interest of $1200 for each of the above-
mentioned times ?
87. Of $100?
88. Of $12?
39. Of $.12 ?
40. Of $120?
41. Of $500?
42. Of $1.20?
43. Of $600?
44. Of $250?
45. Of $.60 ?
179. When to disregard Cents.
If the time is not verj' long, the interest of any sum less
than a dollar can be computed with sufficient accuracy by
referring the principal to the nearest convenient aliquot part
of a dollar.
Thus, the interest of 24 cents, for any ordinary time of calcnlating
interest, will diflFer but a trifle from that of 25 cents, or i" of a dollnr,
for the same time ; the interest of 35 cents will differ but a trifle from
that of 33^ cents, or ^ of a dollar, &c.
1. What is the interest of $599.77 for 9 mo. 17 da. ?
/Sb/tt<ion. — $599.77 = $600 — 23 cents. The interest of $600 for 9
mo. 17 da. (being $3 per month, and 1 dime per day) is $28.70. As 2J
cents is very near 25 cents, its interest must be very near ^ of a cent
per month, which will be about 1 cent for 9 mo. 17 da. This taken
from $28 70 leaves $28.69 as the in^eiest required.
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INTEREST. 259
2. What is the interest of $60.49 for 5 mo. 11 da- ?
8. What is the interest of $59.67 for 8 mo. 13 da. ?
4. What is the interest of $299.51 for 11 mo. 23 da. ?
5 What is the interest of $150.32 for 7 blo. 19 da, ?
6. What is the interest of $6.24 for 9 mo. 5 da. ?
7 Wliat is the interest of $200.42 for 8 mo. 15 da. ?
8. What is the interest of $599.66 for 5 mo. 13 da. ?
9. What is the interest of $119.94 for 13 mo. 7 da. ?
10. What is the mterest of $50.31 for 3 mo. 23 da. ?
180* LUerest at various Raics, obtained from that at 6
per cent.
When the interest is other than 6 per cent per year, we
may first find the interest at 6 per cent, and then take such
part of this as the given rate is of 6 per cent.
Thus, the interest of any sum at 8 per cent is f ^= ^ ss ]^ times its
interest at 6 per cent ; a: 4 j* per cent the interest is -J =» f of the in-
6
terest at 6 per cent, &c.
1. What is the interest of $367.32 for 1 yr. 9 mo. 20 da.,
at 7^ per cent ?
Solution,
a = $367.32 ea principal.
1^ of a = b = 56.732 = int. 20 mo. at 6 per cent.
•j^ of b s= c s=s 3.061 s= int. 1 mo. 20 da. at 6 per cent
b + c = d = 39.793 = int 21 mo. 20 da. at 6 per cent
^ of d = = 9.948 = int. 21 mo. 20 da. at 1^ per cent
d + e = $ 49.741 == int. 1 yr. 9 mo 20 da. at 7j per centai
AfMvoer,
Note. — The work may frequently be facih'tated by obserring that
the interest of any sum at other than 6 per cent is equal to the interest
at 6 p>?r cent of *the same part of that sum that the required rate is of 6
per cent Thus, the interest of any sum for a given time at 3 per cent
is equal to the interest of f , or ^, of that Bum for the same time at 6
per cent. The interest of any sum at 4 j- per cent is eqnal to the iii«e»
est of -~» or f of that sum, at 6 per cent, &c.
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290 INTEREST.
Again. The intereit ot anj sum at other than 6 per cent is eqaal to
its interest at 6 per cent for the same part of the given time that the
required rate is of 6 per cent Thus, the interest of any sum for a
given time at 2 per cent is eqnal to its interest for f , or j-, of the given
nme at 6 per cent.
What is the interest of —
2. $847.38 for 2 yr. 4 mo. 10 da., at 5 per cent ?
8. $483.94 for 3 yr. 5 mo. 26 da., at 7 per cent r
4. $150 for 8 mo. 27 da., at 9 per cent ?
5. $46.38 for 11 mo. 19 da., at 3 per cent?
6. $512.59 for 4 yr. 7 mo. 17 da., at 6^ per cent ?
7. $437.95 from June 17, 1848, to May 19, 1850, at Gf
percent?
8. $978.31 from Jan. 27, 1850, to Sept. 5, 1852, at 5
per cent ?
9. $87.63 from April 21, 1848, to Jan. 7, 1852, at 5}
per cent ?
10. $450 from Aug. 12, 1852, to Oct 7, 1852, at 7^ per
cent?
11. $75 from Nov. 5, 1850, to Jan. 4, 1852, at 3 per
cent?
12. $240 from Sept. 30, 1848, to May 23, 1851, at 1^
per cent ?
181* Interest at various JRates, obtained directly.
Methods similar in character to those illustrated in the fol-
lowing examples and explanations will usually be more brief
than the preceding.
1. What is the interest of $549.84 for 1 yr. 4 mo. 15 da,
at 8 per cent ?
Solution,
a = $549.84 = principal.
«>8 of a Bs b =3 43.987 = int. for 1 yr. at 8 per <;pnt.
j cf b as c = 14.662 = int. for 4 mo. at 8 per cent
^ oi c ss d = 1.833 = int. for 1 5 da. at 8 per cent
b 4- c 4- d = $ 60.482 = int foi 1 yr. 4 mo. 15 da. at 8 per cent ■■
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INTKKEST. 261
beeond Solution. — Since the rate is 8 per cent of the principal, th«
interest for 1^ years, or 15 months, mast bo 1^ times 8 per cent as 10
pel cent == iV of the principal. Hence,
a s=s $549.84 = principaL
:i^ of a = b = 54.984 = int. for 15 mo. at 8 per cent
l\j of b = c = 5.498 = int. for 1 mo. 15 da. at 8 per cent
b + c s= S 60.482 s= int for 16 mo. 15 da. at 8 per cent «■ Am»
2. What is the interest of $537.48 for 3 yr. 8 mo. 15 da.
at 7^ per cent ?
Solution. — The interest foi 2 yr. at 7^- per cent must be 2 timet 7 J,
or i5 per cent of the principal. Hence we have the following writteo
work : —
a =s $537.48 = principaL
.15ofa=sb= 80.622 = interest 2 yr.
j- of b = c =3 40.311 =s interest 1 yr.
^ of a = d =s 26.874 = interest 8 mo.
•j^ of d s=s e =s 1.679 = interest 15 da.
b-f-c + d + e = $149,486 = interest 3 yr. 8 mo. 15 da. at 7^ per
cent
Second Solution. — The interest for 4 yr. at 7^ per cent most be 4
times 7^, or 30 per cent, = -f*^ of the principal. Hence,
a 5=5 $537.48 = principal.
^ of a s b =s 161.244 = interest for 4 yr.
jV of b = c = 10.077 = interest for 3 mo.
J of c = d =3 1.679 = interest for 15 da.
b — c — d =s $149,488 = interest for 3 yr. 8 mo. 15 da.
8. What is the mterest of $537.47 for 1 yr. 10 ma 15 da.
at 3^ per cent ?
Solution. — The interest of any sum for 3 years at 3^ per oeat pet
year will be 10 per cent, or iV of that sum. Hence,
a = $537.47 = principal.
jV of a = h =! 26.873 = int. for 1 yr. 6 mo.
i of b = c = 6.718 = int for 4 mo. 15 da.
b 4- c = $33,591 =: int for 1 jr. lO mo. 15 da.
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S62 INTEREST.
Second Solution,
a ^ $537.47 «= principal.
^ of a ss b = 26.873 = int for 1 yr. 6 mo.
^ of b 8=s c =B 4.479 = int for 3 mo.
j- of c SB d =* 2.239 =3 int for 1 mo. 15 da.
b + c + d = $33,591 = int for 1 yr. 10 mo. 15 d&
Note. — It will be seen that, by this method, we first get the interest
for any convenient time, and then take such part or parts of this as wiD
gire the interest for the required time.
4. What is the interest of $279.04 for 6 yr. 6 mo. 22 da»
at J per cent ?
Solution. — The interest for 4 years at i per cent must be 4 times i,
9r 1 per cent of the principal. Hence,
a = $279.64 = principaL
.01 of a s= b = 2.796 = int. for 4 yr.
^ of b 8=s* c = 1.398 = int for 2 yr.
|^« of b =B d =s .310 = int for 5j mo., or 5 mo. 10 da.
t\r t of c s= e = .023 = int for 12 da.
$4,527 = int for 6 yr. 5 mo. 22 da.
What is the interest of —
5. $483.79 for 3 yr. 3 mo. 18 da. at 4^ per cent?
6. $538.71 for 1 yr. 7 mo. 24 da. at 9 per cent?
7. $875.37 for 2 yr. 7 mo. 13 da. at 8^ per cent ?
8. $63.29 for 5 yr. 11 mo. 10 da. at 8^ per cent ?
Suggestion, — The interest for 6 years, at 8^ per cent, is 50 per cent,
or it of the principal, and since 20 days = -g- of 2 months = tV of I
year = tJ-^ of 6 years, the interest for 20 days must equal -ji^ of the
faiterest for 6 years.
9. What is the interest of $56.84 from March 5, 1850, to
May 25, 1851, at 5 per cent ?
♦ Since 6 j- mo. = ^ of 4 yr., or 48 mo.
t Since 12 da. sr {- of 2 mo., and 2 mo = iV of 2 yr.j 12 da. must
Wnal i of iV, or 1^ of 2 yr.
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INTEREST 263
10. What IS the interest of $138.46 from July 1, 1848, to
Aug. 26, 1852, at 2J per cent?
11. What is the mterest of $273.81 from Sept. 28, 1847,
to Oct. 81, 1852, at 7 per cent ?
12. What is the amount of $783.25 from Aug. 28, 1846,
to Feb. 29, 1852, at 4^ per cent?
13. What is the amount of $57,84 from Jan. 19, 1849, to
Dec. 29, 1852, at 6| per cent ?
14. What is the amount of $278.49, from Sept. 13, 1841,
to Oct. 10, 1846, at 7^ per cent ?
15. Jan. 17, 1850, 1 borrowed 837 dollars, agreeing to pay
interest at the rate of 6 per cent per year, and immediately
put it on interest at the rate of 7jt per cent . Aug. 27, 1852,
I collected the amount due to me, and paid that which I owed.
How much did I gain by the transaction ?
Suggestion. — Since I paid 6 per cent and received 7 J per cent inter-
est on the snm I borrowed, my gain must hare been 1 j- per cent per
year.
16. A merchant, wishing to purchase 9 acres of land at
$378.43 per acre, borrowed money for the purpose at the rjite
of 5 per cent At the end of 3 yr. 9 mo. 15 da. he sold the
land, receiving $400 per acre for ^ of it, and $475.28 per acre
for the remainder. Did he gain or lose, and how much ?
17. Bought 397 yards of cloth at $3.75 per yard, payable
in 6 months, with interest at 7J per cent per year, and imme-
diately sold it for $4 per yard, payable in 6 months, without
interest. When the 6 months had elapsed, I collected the
money due me, and paid my debt. Did I gain or lose, and
how much ?
18. Bought 397 yards of cloth at $4 per yard, payable in
6 months, and inmiediately sold it at $3.75, cash, and put the
money on interest at the rate of 7^ per cent. At the end of
6 months I called in the money I had lent, and paid that
which I owixL Did I gain or lose by the transaction, and
how much ?
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t64 PROMISSORY NOTES.
182. Promissory Notes,
(a.) A PBOMISSORT NOTE, a NOTE OF HAND, OF, OS it IS
more commonlj called, ▲ note, is a written promise to pay a
specified smn of money.
(6.) Annexed if a form of a note, which, with the subsequent expla-
BAtions, will illustrate the principal points connected with this subject.
9^oi i>aCu/e leceu>ea, W kVonru6e to- itcut ^^eolo*
UU6t. ^^<^ ^iOUHV.
(e.) The above note is a promise made bj John Brown to pay
George Smith, or order, one hundred dollars, and is equivalent to the
following: —
Providence, May 1, 1855.
Because of an equivalent value received from George Smith, I
promise to pay to himt or to whomsoever he may order me to pay it,
one hundred dollars, whenever the payment may be demanded of me
by presenting this note ; and also to pay the interest of one hundred
dollars from this date till the time of payment.
John Brown.
(d.) In considering this note, we may observe, —
^irst The " $100 " placed at the left hand upper comer. These fig-
nres do not form an essential part of the note, but are written in order to
enable a person to tell at a glance the amount for which it was given, and
also to guard against any changes which might be made in the body of
the note.
Second. The date, which shows when and where it was written.
Third. The words " value received," which are designed as an ac-
knowledgment that the signer of the note has received an equivalent
from the person to whom it is to be paid.
Fourth. " I promise to pay George Smith, or order," which means
the same as, *' I promise to pay to George Smith, or to whomsoever ho
may order me to pay it."
Fifth. The sum, " one hundred dollars." This should always be
written out in words.
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PROMISSORY NOTES 365
Sixth The phrase " on demaitad," which means whenever he shall
demand payment.
Seventh. The phrase " with interest," which means that interest is
to be paid from the time the note is dated.
Eighth. The signature, " John Brown," which gives validity to the
note, and must be written by Mr. Brown himself, or by some one special-
ly authorized by him.
(e.) The words " value received " are regarded as essential to a
note, it being a principle in law, that no person shall be compelled i»
pay money for which he has not received an equivalent in some form or
other.
(/) The words "or oider" may be omitted, or the words "or
bearer " may be substituted for them. If they are omitted, the note can
only be collected by the person named in it. Such a note is not nego-
tiable. (See 183, a.)
(g.) Some specified time, as "in sixty days," "in three months,"
&c., might be substituted for the phrase " on demand." The meaning
then would be in so many days or months from the date of the note.
Such a note would be called a note on time.
(A.) Notes on time are not regarded as due till three days after the
time specified in the note. Thus, a note payable in sixty days is not
due till the end of sixty-three days. The three days thus added are
called DAYS of qbace.
(t.) If grace is not to be allowed, the form of the note should be,
" in so many days or months without grace."
(j.) If the last day of grace comes on Sunday, or upon a legal holi-
day, as the Fourth of July, Thanksgiving, &c., the note is payable on
the preceding day, and if that be Sunday, or a legal holiday, it is pay-
able on the first day of grace.
(k.) If the phrase " with interest " should be omitted, the note would
not be on interest. If, however, a note on demand is not paid when
the demand is made, or if a note on time is not paid when due, interest
may be afterwards charged, though no mention of it is made in the
note ; so that two notes, one payable in three months, and the other
payable in three months, with interest afterwards, would both be on in-
terest after the expiration of three months. .
(Z.) The form of notes may be varied somewhat without affecting^
their meaning or value. Thus, it makes no difference where the phrases
"value received" and "on demand" are placed, provided they are
abovo the signature. The phrase " to the order of George Smith"
may be substituted for " George Smith or order." These and other
changes will be illustrated in the forms of notes given in subsequent
Dtorts of the book.
23
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J66 NBQOTIADLB NOTL^, INDORSEaiBNTS, 6cc.
{m,) The person whose name appears on a note as promisor if
called the makkb of the note.
(ft.) The person to whom a note is to he paid is called (he ]*atbb or
th) paoMiSBB, or the holder of the note.
(o.) Every note should he written, or at least signed, by the maker
of it, or by some one specially authorized by him. It is then taken b/
the payee, and is his property until it is paid, or until he transfers it to
another. When paid, it should be given up to the signer, who should
Immediately tear off or erase his signature.
(p.) If the maker of a note refuses or neglects to pay it, when the
demand is legally made, at lae proper place and time, the note is said to
be DISHONORED.
(9.) A demand for payment, to be legal,* must be made by actually
presenting the note ; or, if it is payable at a certain place on a given day,
by having the note deposited at that place ready to be presented to the
maker, should he call to pay it
KoTB. — The maker of a note is not obliged to regard a request for
its payment as a legal demand, unless the note be exhibited and ten-
dered to him at the time the request is made ; but if he waives his rignt
to see the note, the demand is legal
183* NegotiaMe Notes^ JBtdorsements, and Protests,
(a.) A NEGOTIABLE NOTE is One that may be transferred
or sold by one person to another.
(6.) Negotiable notes are of two kinds, viz., those payable to a per-
Boiiy or order ^ (as " George Smith, or order, ^) and those payable to a per-
mm, or bearer, (as " George Smith, or hearer,") or simply to " the bearer.^
(c) Those of the second kind are negotiable by mere delivery ; bat
those of the first require a written order of the payee to authorize any
one else to collect the money due on them.
{d.) Such an order is commonly written on the back of the note,
and is called an indorsement.
Thus, if George Smith wishes to transfer the above note to Charles
Woods, he might write on the back of it, " Pay to Charles Woods, or
order. — George Smith."
(«.) This would be an indorsement in full, and would give Charlef
Woods the same title to the note, and the same claim on John Biowc
on account of it that George Smith originally had.
{/.) If the indorsement had been, "Pay to Charles Woods, or
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NEGOTIABLE NOTES, INDORSEMENTS, &C. ^67
bearer " it would give to Charles Woods, or whosoeyer might obtain
legal possession of the note, the full title to it.
The note would then bo negotiable by mere delivery.
ig.) If it had been, " Pay to Charles Woods only," it would give to
Charles Woods the full title to the note, but would prevent his transfer-
ring it to any one else.
(A.) If George Smith had simply written his name on the back ot
the note, it would have been an indorsement equivalent to, ** Pay to the
bearer,*' and would make the note negotiable by mere delivery. Such
nn indorsement is called an indorsement in blank, and is the form
most frequently used.
{i.) By either of the foregoing forms of indorsement, George Smith
would not only authorize some one else to collect the note, but he woula
make himself responsible for several important points. He would
guaranty, —
First. That the note is genuine, and just what it purports to be.
Second. That he has legal possession of it, and a right to transfer it.
Third. That it shall be paid if payment is demanded by presenting
the note to the signer at the proper time.
Fourth. That if not so paid, he will pay it himself, if properly
notified.
If he does not wish to guaranty the last two points, he writes the
\iH)rd8 "without recourse" before his name. The indorsement in this
form is a guaranty of the fiist two points, but not of the last two.
(/) Any person may write his name as a special indorser on a note
which he does not own. Such an indorsement would not aflFect the
negotiability of a note, but it would make the indorser responsible for
its payment, in case the maker should not pay it. Indeed, it might im-
pose upon him all the obligations with regard to the note, which rest
upon the original maker.
{Jc.) A person who indorses a note under any circumstances incurs
all the obligations of such an indorsement, even though he may be
ignorant of them at the time.*
(/.) If a note Is indorsed by several persons, each of them makei
himself responsible for these points to whoever may afterwards get legal
possession of it.
(w.) A person to whom an indorsed note is transferred, by the mere
act of receiving it, agrees with all the indorsers. except the special
ones, and under some circumstances with them, —
♦ Many a man has been redufed from affluence to poverty, by merely
writing his name on the back of a note "just to accommodate a friend."
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268 NEGOTIABLE NOTES, IXDORSEMUNTS, &0.
First. That payment shall be demanded of the signer within a return
aWe titne^ if the note is payable on demand ; and :n the very day it becoma
due^ if it be on time.
Second. That he will not consent to any delay in the time of pay-
ment.
Third. That if the note is dishonored he will at once get it protest-
cd, (seo (u.) (i7.) p. 269,) or, on that day or the next, inform the indorser
that it is dishonored, and notify him that Le will be held responsible
for its payment This information should be given by telling him per-
nonally, or by leaving a letter at his place of business or dwelling, or by
mailing one to his address.
(n.) If the holder of a note fails to comply with any one of these
conditions, he releases every indorser except the special ones, and some-
times even them, from all obligation to pay it ; but he does not in any
way aflfect the obligation of the original promisor.
(o.) The question. What is a reasonable time in the case of a note
payable on demand 1 must depend for its answer somewhat on circuni'
stances.
In Massachusetts, the indorser of a note payable on demand is by
statute excused, if payment be not demanded within sixty days from
the date of the note.
{p.) If no particular place for payment is specified in the note, it
may be presented at the signer's counting house or place of business, in
business hours, or at his dwelling house, at a time when he may reason-
ably be supposed to be at home, and not to have gone to bed.
(q.) If it is payable at some specified place, as at a bank or at some
counting room, the demand must be made at that place. In case the
promisor does not appear there to pay the note on the day it falls due,
it is dishonored, and notice should be sent to the indorsers, as though
payment had been demanded and refused.
(r.) When there are several indorsers to the note, notice that it is dis-
honored should be sent to each whom the possessor of the note wishes
to hold responsible for its payment. He may collect the money of either,
or, if he resorts to legal measures, may commence suits against any of
them, or against all at once. When, however, he collects the money of
any one, his demands against the others are void.
(s.) Each indorser may require any one whose name precedes his
•wn to make good to him the loss he may sustain on account of the
note, provided he gives notice of his intention to do so the day that he
receives his own notice, or the day after.
For instance, suppose that the foregoing note comes to me indorsed
by George Smith, Charles Woods, Silas Bacon, and Edward Jones, and
^ tliat it is dishonored. I shall immediately inform Smith, Woods, BacoUi
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JOINT AND SEVERAL NOTES. 2G9
and Jones of the fact, and tell each that I sliall look to him for payment
Jones will on the same day, or day after receiving the notice, inform
Smith, Woods, and Bacon of it, and also that he expects them to make
good to him any money he may have to pay on account of it . Bacon
would write a similar letter to Smith and Woods, and Woods would write
one to Smith ; and if Smith should be obliged to pay it, he could hold
only the signer of the note responsible.
(t) A PROTEST is a formal declaration, made by an officer called a
Notary P-.blic, that a note is dishonored, and that the indorser will be
hcM responsible for it.
(«.) The common method of protesting a note is for the holder to
take it to a notary public, and get him to demand payment of the
promisor. If the demand be refused, the notary makes out a protest,
i. e^ he writes (usually on a copy of the note) a foimal declaration that
it is dishonored, and sends a copy of it to each indorser. The protest
should be made out on the very day the note is due ; otherwise it has no
value.
{v.) Although a notice sent by the holder of the note is usually
regarded as sufficient to make the indorsci-s liable for its payment, it is a
safer course to have a formal protest made out.
184. Joi7it and Several Notes.
(a.) Notes are sometimes signed by more than one person. If tho
note is so worded as to show that the signers are together responsible
for its payment, it is a "joint note ; " but if so worded as to show that
tlie signers together and separately are responsible for its payment, it is
a "joint and several note."
(6.) There seems to be but little difference between a joint note and
a joint and several note, as far as regards tlie liabilities of the promisors.
Any signer to either may be compelled to pay its whole amount. The
chief difference between them is, that an action commenced to enforce
payment of a joint note must be commenced against all the promisors
jointly, while an action commenced to enforce payment of a joint and
several note may be commenced against them jointly or individually.
(c.) If judgment is given in favor of the holder of a joint note, he
may compel either promisor to pay its full amount, as much as if it had
been a joint and several note. One who is thus compelled to pay the
amount of the note may recover from each of the other promisors the
share which each ought justly to pay. In a joint and several note, or «
several note, he may or may not recover from the other promisors, ac-
eording to the circumstances under which the note was gi-on.
28*
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270 RENEWAL OF NOTES.
050. "^Piouaervee, §ft)a^ f, 1^55.
(J^oi ooAju, iecetueo, u>e Mnniuif kioriu.^ bo- Kcl<^
^ >um«S> ^HoA^ o^> oioei, KAa dou!a^, ow otnuurvdy unia
(i.) The abore is a joint note, because it is the promise of Spragae
and Hale to pay jointly, or together, a specified sum of money.
(«.) The holder of the note may demand payment of either Sprague
or Hale ; but if he wishes to hold any indorser or indorsers responsible,
he must demand payment of both. If he commences an action on ac
count of it, he must commence it against both jointly.
(/) When a note is signed by one person as principal, and another
as surety, the holder of it may demand payment of, and commence
action against, the surety instead of the principal, if for any reason ho
chooses so to do.
18t5. Renewal of Notes,
(a.) If the holder of a note allows it to run for six years without col-
bcjUng any thing on it, or otherwise getting it renewed, it becomes out-
lawed, and he cannot afterwards compel the promisor to pay it.
(6.) A note under seal, and in some states (as Massachusetts and
Connecticut) an attested note, is not outlawed under twenty years.
(c.) Any act by which the promisor acknowledges the validity of a
note renews it. Care should be taken to have the renewal made in the
presence of witnesses, or to have evidence of it in the handwriting of
the promisor. For instance, —
{d.) When a part of the money due on a note is paid, a receipt for it
should be written on the note. Such a receipt is called an indorsement
rims, if twelve dollars be received on some note, the following would
be written: "Jan. 1, 1855, received twelve dollars." This indorsement
should be written by the one who pays the money, as it is equivalent to
a renewal of the note, and it may afterwards be important for the holder
to prove that the money was really paid at that time on account of
the note.
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FOEMS OP NOTES. 271
186. Exercises.
0500. "Boi^., ©Mdo^ T, i^55.
Qnuneav dcui6- optet dote, Q) kVomt^ ka kcM
^ortve^' ^Veuv, o^ »Vaei, fu>e Uiutdiea aoCui/i^. ^Uai-ae !€•
celuea. uorvwi/cC >oluvi^&rv.
Who is the maker of the above note ? To whom docs he promise to
pay it 1 Who shall take possession of the note ? Can he transfer it 1
If so, how ? What form of indorsement would be used if he wished to
transfer it to Francis Baker, or his order? To Francis Baker only?"
To Francis Baker, or whoever should have possession of it 1 To who-
ever should have possession of it without naming any person 1 How
should it be indorsed in each of the above cases, if the indorser does not
wish to make himself responsible for its payment ? Is it on interest 1 On
what month, and what day of the month, will the note become due 1
If this note should be indorsed first by James Drew, then by Frz^acis
Baker, and then by William Davis, and then Charles Morton should
come into legal possession of it, of whom and when ought he to de-
mand payment? If pajrment is refused, what ought he to do ? What
ought each indorser to do on receiving notice that the note is dishon-
ored ? Who should take possession of it when it is finally paid ? What
disposition should be made of it ? What effect would it have on the
negotiaoility of the above note, if the words " or order " were omitted ?
What if the words " or bearer " were substituted ?
0500. '^iooLSeace, 9<6&, f, 1^55.
Vth acm^ttrvo. uyMv irvte1e6b aitei tn/Ve-e n^<M^tn/6^, Q)
|i/tom.t^ to hoAL to llie otaeV op iJWen'ui ^'uuv fLue luwta^co
ooUa/V6. ^Uau»/e Vccemeo.
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97% BANKS AND BANKING.
(2.)
$1000.
{»•)
cUvoE, orta J'ajnvt^ j^OAM^otVy oA iuAth^ jviorm/^ to- Ivcul
StkiuaVo «7«^!eea eiAnfc Kitrtuieo ooCuti^, orv ael1^arl/a, uhuI uv-
xonnfwli TxwKijo.Yu
Let the scholar write each of the above notes and explain their mean-
injr, and the meaning of all their points ; let him also change their form,
and indorse them in varioos ways.
187. Baiiks and Banking.
(a.) A bank is an institution or corporation for the pur-
pose of trafficking in money.
(b.) Banks usually receive money on deposit, loan money on interest,
s^and issue bank notes, or bank bills, i. e., notes payable in specie to ihe
bearer on demand at the bank, and intended to circulate as money.
(c) When money is loaned by a bank, it is commonly
made payable at the end of a given number of days, and the
jnftTett for that time and the three days of grace is deducted
at the time it is borrowed.
TIius, on a not€ of $500, payable in SO days, I shall receive at a bank
$f.OO, miuus the interest of $500 for n days, i. o , $500 — $2.75 = $497.25.
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BANKS AND BANKING. 273
((I) By this arrangement the banks receive interest on a
larger sum of money than they lend.
Thns, in the above exam pi 3, the bank receives interest on $500, while
the sum actually lent is only $497.25.
(e.) Bank interest is called discount, because it is thus
deducted from the face of the note, i. e., from the sum for
which the note is given.
(/.) The note on which the money is received is said to
be discounted.
(g.) To present a practical illustration of this subject, we will sup-
pose the following case : —
On May 9, 1855, George Guild, being in want of money, wrote a
note promising to pay at the Merchants' Bank, to the order of Alfred
Hall, $600, in 60 days, and got Mr, Hall to indorse it. He then applied
to the officers of the bank to discount it, and they decided to do so. Ho
forthwith presented the note to the cashier, who deducted the interest of
$600 for 63 days, and paid him the balance, $593.70. Guild took the
money, and liad the use of it till July 11, when the i>ote became due.
He then paid to the cashier of the bank the $600 due on the note, and
the transaction was settled.
(A.) By considering the above, it will be seen that Guild paid to the
bank the $593.70 which he had borrowed, together with the interest of
$600; so that he paid the interest of $6.30 (i. e., of the bank discount)
more than he liad the use of.
(t.) If he had wished to keep the money as much longer, he would
on the last day of grace have written a new note, differing from the
former only in the date, and have got it indorsed as before. As this
new note would be worth $593.70 at the bank, he could by giving it, and
$6.30 besides, to the cashier, pay the amount due at the bank. At the
end of 63 days he would again owe the bank $600.
(j.) Now, it is obvious that during all this time he has been paying
the interest of $600, while he has had the use of but $593.70, and that
therefore he has paid the interest of $6.30 more than he has used. Be-
Bidcs this he loses the use for 63 days of the $6.30 he paid on renewing
the note. Hence, as the use of a sum is worth its interest, he virtually
pays the interest of $6.30 more than he receives for 126 days -f- 63 days,
or 189 days.
[k.) If Mr. Guild should fail to appear at the bank to pay the note
before the close of bank hours on the last day of grace, the note would
be protested, and notice sent by a notary public to Mr. Hall, who would
then be held responsible for its payment
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274 BANKS AND BANKIKO.
1. A note of $1200, payable in 60 days, was discounted at
a bank at 6 per cent. How much was received on it ?
Schitioa, — The interest of $1200 for 63 days, being 2 dimes per diiy
h tI2.60, which, deducted from $1200, leaves $1187.40 as the sum re-
ceived.
2. How much would be received at a bank on a note of
$200, payable in 90 days ?
3. How much would be received at a bank on a note of
$360, payable in 30 days ?
4. I got my note for $1000, payable in 90 days, discounted
at a bank, and immediately put the money received on it at
interest. When the note became due, I collected the amount
of what I had put on interest, and paid my note at the bank.
How much did I lose by the ti*ansaction ? How does the sum
lost compare with the interest of the bank discount for the
given time ?
5. My note for $1000, payable in 6 months, was discount-
ed at a bank, and I immediately put the money received on it
at interest. When the note became due, I collected the sum
due me, and paid that which I owed at the bank. How much
did I lose by the transaction ?
' 6. I had my note for $500, payable in 2 months, discount-
ed at a bank, and immediately put the money on interest.
When the note became due, I renewed it for the same time as
before ; and when the new note became due, I collected the
amount due me, and paid my note at the bank. How much
did I lose ?
Suggestions. — F^m a consideration of/flie methods of reckoning in-
terest at banks, it is\yident that fronyfne time the first note was dis-
counted to the time tu^ second was/paid, I paid interest on the bank
discount more than I recced, and^at at the end of two months three
days I paid a sum equal to ihe bank discount. Hence, I lost the inter-
est of the bank discount for ^^o. 6 da., plus 2 mo. 3 da., = 6 mo. ^ da.
Or, since I paid nothing^t the ^nk, except the bank discount at the
time of renewing the note, and the second note when it became due,
the actual value, at the time of settlement, of the sums paid, will be the
vmcunt of the bank discount for 2 mo. 3 da., plus the face of the note.
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ENGLISH METHOD OP COMPUTING INTEREST. 275
The sum received will be the amount for 4 mo. 6 da. of the mrney ob
taincd at the bank on the first note. The difi^erence between the value.«f
paid and received is the loss.
^^^4>yf 7. I had my note for $600, payable in 4 monlhs, discounted
at a bank, and immediately lent the money received on it fcr
just one year. When my note at the bank became due, I re-
newed it for the same time as before, and when this new note
became due, I renewed it for such time that it became due at
the end of the year, when I collected the amount of the sum
I had lent, and paid my note at the bank. How much did I
lose by the transactions ?
188. English Method of computing Interest
(a.) In. England, time is reckoned in years and days, but
never in months. The year is regarded as 365 days. Inter-
est is usually computed by first finding it for the years, and
then for the days.
(5.) In computing it for the days, it is well to notice that 73 days =s
"t of 1 year, that 5 days =! 7^ of 1 year, and that 1 day = ^^-y of 1
year.
(c.) When any part of the principal'is expressed^ in shil-
lings, pence, and farthings, it should be reduced to the decimal
of a pound.*
* Probably the simplest method of doing this is to regard each shilling
as ^, or .05 of £1, and each farthing as sirTf or .OOI2V of £1. We shall
then have as many times .05 of £1 as there are shillings, plus as many
times .0012^ of £1 as there are farthings in the pence and farthings. But
as all values less than i of .001 of £1 are so small that they may be dis-
regarded, the result will be sufficiently accurate for ordinary purposes, if
we regard each farthing as .001 of £1, observing io add .001 if there are
more than 12 and less than 36 farthings, and .002 if there are more than
36. By adding this result to the value of the shillings, we shall have the
decimal expression required. For example : To find what part of £1 is
equal to 9 s. 8 d. 1 qr., we have 9 s. = 9 times £.05 = £.45 ; 8 d. 1 qr. =s
33 qr. = £.033 -f £.001 = £.034 Therefore, 9 s. 8 d. 1 qr. = £.45 + £.034
«r £.484 The reverse operation will get the value of the decimal expre»»
t'on, in terms of shillings, pence, and farthings.
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S(75 ENGLISH METHOD OF COHPDTINa INTBRBST.
1. What is tho interest of £327 17 s. 7 d. from Majr 7,
1851, to Sept. 4, 1852, at 5 per cent?
Solution. — From May 7, 1851, to May 7, 1852, is 1 year. There mr«
84 days left in May, to which adding the 30 in June, the 31 in Julj, the
81 in Aag:u8t, and the 4 in September, gires 120 days. The time, then.
If 1 year 120 days. Tho principal equals £327.879 Hence we bars
the following written work : —
a ^ j£327.879 s= principaL
i)5, or aV of a «= b = 16.39395 = int for 1 yr.
r^ of b = c = .044914 = int for I da.
120 times c s d ss 5.389680 = int for 120 da.
b + d = e = £21.78363 = int for 1 yr. 120 da.
Or we may have the following : —
a = £327.879 = principaL
.05, or sV of a = b = 16.39395 s= int for 1 yr.
tV of b = c = .224574 = int for 5 da.
23 times c s= d as 5.165202 = int for 115 da.
b + c + d = e= £21.783726 = int for 1 yr. 120 da.
Calling this £21.784, we hare £21 15 s. 8d. 1 qr. as the answer.
The multiplications required in solving these examples render it neces*
sary to carry out the work to places below thousandths, though we do
not care to have them appear in the answer.
2. What is the interest of £47 9 8. 4 d. 1 qr. from May
17. 1849, to Aug. 23, 1852, at 5 per cent?
3. What is the interest of £148 19 s. 9 d. 3 qr. from Oct.
23. 1850, to Nov. 11, 1852, at 5 per cent ?
4. What is the amount of £361 13 s. 2d. 1 qr. from Jolj
18, 1847, to April 12, 1850, at 5 per cent?
5. What is the amount of £248 18 s. 10 d. 3 qr. from
Dec 5, 1849, to March 3, 1852, at ^ per cent?
6. What is the interest of £548 15 s. 7 d. 3 qr. from Jolj
29, 1847, to March 12, 1850, at 2J per cent?
7. What is the amount of <£258 19 s. 5 d. 2 qr. from Jan.
I, 1849, to Sept 29, 1852, at 4 per cent?
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PARTIAL PAYMENTS. 277
8. What is the amount of £329 7 s. Id. 3 qr. from Nov,
13, 1850, to Dec 1, 1852, at 3 per cent ?
9. What is the interest of £481 13 s. 5 d. 1 qr. from
April 19, 1842, to May 3, 1847, at 5 per cent?
10. What is the amount of £222 2 s. 2 d. 2 qr. from Feb.
29, 1848, to Jan. 1, 1852, at 4 per cent?
189« Partial Payments.
(a.) The principle adopted by the Supreme Court of the
United States, and by that of Massachusetts and most of the
other states, as the one to be applied in determining the sura
due on a promissory note or bond on which payments have
been made at different times, is, that as much of the payment
as is necessary to pay the interest due at the time the pay-
ment is made should be appropriated to that purpose, and the
surplus to the payment of the principal. The balance then
due will form a new principal on interest as was the original
principal. If, however, any payment is less than the interest
at the time due, the principal remains unaltered, and on inter-
est, till some payment is made, which, with the preceding
neglected payments, is more than sufficient to pay the inter-
est ; when we proceed as if a single payment, equal to the
sum of the last payment and the preceding neglected ones had
been made.*
EXAMPLE.
$750.00 Boston, April 7, 1848.
For value received, I promise to pay James Sullivan, or
order, seven hundred and fifty dollars, on demand, with in-
terest. Edwai'd Delano.
* The method adopted by tho court of Connecticut differs from the
above only in thb respect — that if a payment greater than the interest at
the time due be made before the principal has been on interest one year,
the person making it is allowed interest on it to the end of the year ; that
iSf its amount from the time it was made to the end of the year, is deduct-
ed from the amount of the principal to the same time. If settlement De
made before the principal has been on inteiest one year, interest is allowea
on the payments from the time thev were made to the time of settlement.
24
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278 PAUTIAL PAniENTS.
On this note are the following indorsements : —
Jan. 17, 1849. Received one hundred dollars.
3Urch 13, 1850. Received tweotj-five dollara.
Feb. 19, 1851. Received thirty dollars.
Ang. 3, 1851. Received two hundred dollars.
Jan. 1, 1852. Received one hundred and fifty dollars.
What was due on the note at the time of settlement^ Aug.
14,1852?
(6.) The following exhibits a good form of writing the work in soch
txainples, and, in connection with the explanations following, will be •
•uflicient illostration of the process : —
$750.00 = 1st principal, April 7, 1848. -
35.00 s=s mtjn(^aysji[ l^ dimes per day.
$785.00 = a m > i dm Jan . 17^ 104 9 .
$100.00 ss 1st payment.
$685.00 = balance due Jan. 17, 1849.
6^.50/= int 20 mo.
34^ =s int. 10 mo.
I =: int. 15 da.
: int 2 da.
2 yr. 6 mo. 17 da.
$789.69 = amU llUB AUg. y, liilil.
255.00 = 2d, 3d, and 4th payments.
$534.69 = bal. due Aug. 3, 1851.
13.367-^= int. 5 mo. 1 , , . ^^ _
.089==intlda. | 5 mo. - 1 da, or 4 mo. 29 da.
$547,968 = Kmti due Jau. 1, 18&2I
1 50.00 s= 5th payment
$397,968 = bal. due Jan. 1, 1852.
1,3.265 =3 int. 6 mo. 20 da. ]
i^6 = int. 20 da. V 7 mo. 13 da.
/.I9e = int. 3 da. J
' $412,757 = amt. due Aug. 14, 1 852, = Am,
Explanation. — As it is obvious that the first payment was greater
than the interest at the time due, we get the amount of the note to that
time, and deduct from it the payment. The remainder is a new princi-
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PARTIAL PAITMENTS. 279
pal Tne second payment, $25, is evidently less than the interest then
due ,• for the time is ot er one year, while the principal, between $600
and $700, gives more than $3G intcretit per year. Similar considera
tions will at once shoB that the interest to the time of the third pay
ment must be greater tlian the second and third payments together.
But the fourth payment, together with the second and third, is very
evidently more than sufficient to pay the interest then due j therefore we
get the amount of the new principal to the time of the fourth payment,
and subtract from it the sum of the second, third, and fourth payments,
thus getting our third principal. As it is evident that the interest of
this principal to the time of the fifth payment is less than that payment,
we find its amount, and from it subtract the payment. This gives ua
the fourth principal, which is on interest till the time of settlement ; and
hence its amount is the sum due.
(c.) The above is really equal to the following simple problems, each
of which is very easy of solution : —
Is the interest of $750 from April 7, 1848, to Jan. 17, 1849, more or
less than $100, the first payment ? What, then, is the amount of $750 for
that time 1 How much will be due after paying the $100 ? Is the in-
terest of this to March 13, 1850, more or less than the $25 at that time
paid ? Is the interest to Feb. 19, 1851, greater or less than $25 -f- $30,
or $55, the sum of the second and third payments ? Is the interest from
Jan. 17, 1849, to Aug. 3, 1851, greater or less than $255, the sum of the
second, third, and fourth payments ? What, then, is the amount of
$685 from Jan. 17, 1849, to Aug. 3, 1851 1 How much will be due after
deducting the $255 ? Is the interest of this from Aug. 3, 1851, to Jan.
1, 1852, greater or less than $150? What, then, is the amount of
$534.69 for that time? How much will be due after deducting $150,
the fifth payment ? What is the amount of this from Jan. 1, 1852, to
Aug. 14, 1852 ? Wliat, then, was due Aug. 14, 1852 ?
Every problem in partial payments can be resolved into simple ones ;
and if the pupil will use a little care in determining what these are, and
be sure that he performs each correctly, he may obtain a true result the
first time of performing the work. Nothing short of this should satisfy
him.
2. $850.00 Bostou, April 7, 1847.
For value received, I promise to pay wAIbert Simmons, or
Older, eight hundred and fifty dollars, on demand, with in*
terest Isaac Goodrich.
On thife note were the following indorsements : —
J ae 19, 1848. Received one hundred and twenty- five dollars
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t60 PARTIAL PATMENTB*
Jan. 7, 1849. Received eighty-three dollars.
Bept 27, 1849. Received oqo haodred doUart.
May 1, 1850. Received twenty dollars.
Aug. 28, 1850. Received two hondred dollars.
Jan. 1, 1851. Received one hundred dollars.
Ilow much was due on the note Oct. 13, 1851 ?
8. SIOOO.OO Providence, Nov. 28, 1848.
I piomise to paj Bradford Allen, or order, one thousand
dollars, on demand, with interest. Value received.
Henry Williams.
On this note were the following indorsements : —
July 23, 1849. Received eighty dollars.
Feb. 28, 1850. Received fifteen dollars.
June 27, 1850. Received twenty dollars.
April 2, 1851. Received twenty-five dollars.
Dec. 20, 1851. Received five hundred dollars.
May 17, 1852. Received three hundred dollars.
How much was due Aug. 14, 1852 ?
Mf^f 7
4. $645tV^ Worcester, Dec. 20, 1846.
For value received, we promise to pay Alfred Lincoln, or
order, six hundred and forty-five dollars and seventy-fiv<%
cents, in three months, with interest after.
Thompson & French. ^
Indorsements : —
Nov. 8, 1848. Received forty dollars.
April 16, 1849. Received three hundred dollars.
March 10, 1851. Received two hundred and fifty dollars.
Sept. 8, 1851. Received sixty dollars.
How much was due Jan. 1, 1852 ?
5. $1275.00 Bridgewater, Sept.'29, 1845.
For value received, we promise to pay Lincoln and Wood
twelve hundred and seventy- five dollars, on demand, with in-
terest. Paine, Root, & Co.
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PAKTIAL PAYMENTS. 281
(iidorsements : —
Aug. 5, 1846. Received three hundred dollars.
Sept. 22, 1847. Received four hundred dollars.
May 25, 1848. Received two hundred dollars.
June 17. 1849. Received one hundred and fifty dollars.
Nov. 13, 1850 Received one hundred and fifty dollars.
What was the balance due March 1, 1851 ?
6. $3000.00 Lowell, April 3, 1849.
For value received, I promise to pay the order of James
Wyman three thousand dollars, on demand, with interest
%fler four months. t^,, ji -o i..
Edward Kobmson.
Attest, George Stone.
Indorsements : —
Nov. 1, 1849. Received five hundred dollars.
Dec. 27, 1850. Received ninety dollars.
March 25, 1851. R^eceived fifty dollars.
July 18, 1851. Received six hundred dollars.
Sept 13, 1851. Received one thousand dollars.
Jan. 1, 1852. Received one thousand dollars.
The note was settled Nov. 8, 1852. How much was due ?
190 • Merchants^ Method when Debts are paid within a
Tear.
(a.) When notes or debts of any kind, on which partial
payments have been made, are paid in full within one year
from the time interest commences, merchants often determine
the sum to be paid on settlement, as they would if nothing is
due on a note till it is paid in full ; that is, they find the
amount of the note to the time of settlement, and the amount
of each payment from the time it was made till the time of
settlement, and then consider the excess of the amount of the
note over the sum of the amounts of the several payments to
be the sura due on settlement.
24 ♦
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182 PARTIAL PATMFNT8.
1. $500.()0 Worcester, July 8, 1851.
For value received, I promise to pay John F. Barnard, ox
order, five hundred dollars, on demand, with interest.
William H. West.
Indorsements : —
Sept 23, 1851. KccciTed sixty dollars.
Kov. 20, 1851. Receired one hundred dollars.
Jan. 17, 1852. Received two hundred dollars.
Feb. 8, 1852. Received fifty dollars.
How much was due May 11, 1852 ?
SotUion,
$500.00 as= principal.
25.25 s=3 int. 10 mo. 3 da.
$525.25 = amt. of note to May 11, 1852.
$ 60.00 = let payment, Sept. 23, 1851.
2.28 = int. 7 mo. 18 da,
100.00 = 2d payment, Nor. 20, 1851.
2.85 t=4nt. 5 mo. 21 da.
200.00 = 3d payment, Jan. 17, 1852.
3.80 «= int. 3 mo. 24 da.
50.00 s= 4th payment, Feb. 8, 1852
.775 s= int 3 mo. 3 da.
$419,705 = amt. of payments, May 11, 1852.
$106.55 = bnl. due May 11, 1852, = Ans.
2. $728.00 Springfield, Sept 7, 1849.
For value received, I promise to pay A. Parish, or order,
seven hundred and twenty-eight dollars, on demand, with in.
lerest. William Mitchell.
Indorsements : —
Oct. 3, 1849. Received eighty dollars.
Dec. I, 1849. Received ninety dollars.
Feb. 4, 1850. Received one hundred dollars
March 2, 1850. Received forty, dollars.
June 1, 1850. Received eighty dollar.
How much was due Aug. 2, 1850 ?
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PARTIAL PAYMENTS. 283
8, $583^^^ Lowell, Jan. 18, 1850.
For value received, I promise to pay C. C. Chase, or order,
five hundred and eighty-thi*ee dollars and seventy-five centS)
^ on demand, with interest. A. H. Fiske.
Indorsements : — »
March 5, 1850. Received fifty dollars.
April 1, 1850. Received seventy-five dollars.
June 17, ISSC^. Received twenty-eight dollars.
July 3, 1850. Received one hundred doHars.
Oct 2, 1850. Received ninety dollars.
Dec 27, 1850. Received seventy dollars.
How much was due Jan. 7, 1851 ?
191. To find the Time.
The methods illustrated in the following solutions will ena-
ble us to find the time, when we know the principal, interest,
and rate.
1. How long must $420 be on interest at 6 per cent to
gain $32.27 ?
iSb/utton.— The interest of $420 for 6 days is $.42 If it takes 6 days
to gain $.42, it will take ^V o^ 6 days to gain $.01, and 3227 times the
last result to gain $32.27
461
3ff I of 6 days = ^ ~ days = 461 days = 15 mo.
^^ [11 days.
2. How long must $357 be on interest at 6 per cent to
gain $29,869 ?
Solution. — Tlie interest of $357 for 6 days is $.357 If it takes 6
days to gain $.357, it will take -§f f ^ of 6 days to gain $29,869
251 2
t|||ft of 6 davs = ???^fL2Lr days = 502 days = 16 mo
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J84 EQUATION OP PAYMENTS.
8. How long must $186^0 be on interest at 7 per oeni ta
gain $2,798 ?
Sotuiiotu — The interest of $136.80 for I year, or 360 days, at 7 per
cent, is $9^76. If it takes 360 days to gain $9,576, it will take |if|
of 360 days to gain $2,793
21 5
fill of 860 da. = ^^^^^ ^^ da. = 105 da. = 3 mo.
KoTB. — By this method of solution, we first select some convenient
time for which to compute the interest Then the required time will be
the same part of the time selected, that the given interest is of the inter-
est for the selected time. The selected time should be one for which the
interest can be easily computed, as, when the rate is 6 per cent per year,
6 do., 60 da. or 2 mo., 600 da. or 20 mo., 6000 da. or 200 mo. ; and
when the rate is other than 6 per cent, 1 year = 360 da., or such part
of 1 year as shall give for the interest 1 per cent, or some other equally
convenient part of the principal.
When there is a fractional part of a day in the result, the fraction
may be omitted if it be less than ^ ; but if it be more than ^, 1 may bo
added to the number of days.
In how long time at interest will —
4. $427.32 gain $19.68 at 6 per cent?
5. $186.75 gain $12.45 at 6 per cent ?
6. $378.50 gain $4,542 at 7 per cent?
7. $56.34 gain $18.78 at 8 percent?
8. $873.70 gain $17,474 at 5 per cent?
9. $594.00 gain $60,654 at 4 per cent ?
193. Equation of Payments.
(a.) When one man owes another sums of money payable
at different times, it may be desirable to determine when the
whole can be paid without gain or loss to either party. The
process of doing this is called equation of payments, and
the time sought is called the equated time.
(h,) It is obvious that if a debt be not paid till after it has
becomi? due, the debtor gains the use o( it from the time it
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EQUATION OF PAYMENTS. 285
became due to the time of payment ; while if it he paid be-
fore it becomes due, the debtor loses the use of the sum paid
from the time of payment to the time when it would justly
liave been due. The use of any sum of money is regarded
as worth its interest for the time it is used.
(c) The application of the foregoing principles is Ulos*
trated in the following problems and solutions : —
1. Mr. Lincoln owes Mr. Wood $400, due in 5 months,
$600, due in 9 months, and $200, due in 12 months. When
can the whole be paid without gain or loss to either party ?
Solution. — By the conditions of the question, Mr. Lincoln is en-
titled to the use or
Interest of $400 for 5 mo. ==• $10.00
Interest of $600 for 9 mo. = $27.00
Interest of $200 for 12 mo. = $12.00
Or to use $1200 till its int = $49.00
The interest of $1200 being $6 per month, he is entitled to keep it m
many months as there are times $6 in $49, which are 8^ times. There-
fore he is entitled to keep it 8-^ months, or 8 months and 5 days.
Ftrtt Proof. — By paying the whole at the equated time, Mr. Lincolu
gains the use of the first debt from the time it was due to the equated
time, and loses that of the second and third from the equated time to
the time when they would otherwise have been due. That is, he
gains interest of $400.00 for 3 mo. 5 da. = $6.33^
and loses interest of $600.00 for 25 da. = $2.50
and loses interest of $200.00 for 3 mo. 25 da. = $3.83 j
Sum of losses == $6.33^ = the gain,
which shows the work to be correct.
Second Proof. — If each debt should be paid when it becomes due,
Mr. Wood will, when the last debt is paid, have had the use of $400 for
7 mo. and of $600 for 3 mo., which at 6 per cent is equivalent to $14 +
$9 =3 $23 interest. If, however, the sum of the debts should be paid at
the equated time, Mr. Wood would, at the end of 12 months, when the
last debt would otherwise have been paid, have had the use of $1200 for
3 mo. 25 da., which, at 6 per cent, is worth $23 interest This shows
that he would have the same interest in one case as in the other, and
thus proves the first result correct.
Second Solution . — Another solution similar in character to the last
can be obtaiued by ascertaining how much would be gained or lost by
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28C KQCATION OF PATMEKTS,
pftjiog the entire debt at ao j assameil time, and froci 'hmt getting tbf
•quted time.
For instance, sappo^ that the entire debt had been paid at the end
of 9 months. Then Mr. Lincoln woald have
I gained interest on $400 foi 4 months =: $8.00
and lost interest on $200 for 3 months &= $3.00
equivalent to a gain of $5X)0
which shows that 9 months is as many days longer than the trne time
as it will uke for $1200 to gain $5 at interest We find (by 191) that
it will take $1200, at 6 per cent interest, 25 days to gain $5. Therefore
the equated time » 9 mo. — 25 da. s= 8 mo. 5 da.
Third Solution. — When the nnmbers are convenient, as in this ex
ample, a method like the following can be used to advantage : —
The sum of the debts is $1200, of which the first debt is j>, the
•econd j-, and the third ^. But the use of ^ of a sum 5 mo. is worth
as much as the use of the whole of it for ^ of 5 mo., or 1§ mo. ; the
use of j- of a sura for 9 mo. is worth as much as the use of the whole
of it for j- of 9 mo., or 4^ mo. ; and the use of ^ of a sum for 12 mo. is
worth as much as the use of the whole of it for ^ of 12 mo., or 2 mo.
Therefore Mr. Lincoln is entitled to the use of the sum of the debts for
1§ mo. -f- 4^ mo. + 2 mo. = 8^ mo. == 8 mo. 5 da.
Fourth Solution. — The following method is much used, but we think
the method by interest will ordinarily be found preferable : —
The use of $400 for 5 rao. is worth as much as the use of $1 for 400
times 5 mo., or 2000 mo. The use of $600 for 9 mo. is worth as much
as the use of $1 for 600 times 9 mo., or 5400 mo. The use of $200 for
12 mo. is worth as much as the use of $1 for 200 times 12 mo., or 2400
mo. Therefore Mr. Lincoln is entitled to the use of the entire debt for
such time as will be equivalent to the use of $1 for 2000 mo. -f- 5400
mo. + 2400 mo., or 9800 mo. But as the use of $1 for 9800 mo. is
equivalent to the use of $1200 for oW of 9800 mo., or Si mo., he can
keep the entire debt 8^- mo., or 8 mo. 5 da.
Notes. — First As the equated time will be the same, whatever be
the rate of interest, the rate may be considered to be that which is most
easily calculated.
Second. The equated time will frequently contain a fraction of a
day; but if the fraction be less than J-, it may be disregarded, or if it
be more than ]}■, 1 may be added to the number of days.
2. A owes B $250, due in 3 mo., $400, due in 6 mo., and
$350, due in 8 mo. What is the equated time of payment ?
3. I owe $700, payable as follows : $150 in 3 mo., $184
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TO FIND DATE OF EQUATED TIME. 287
in 7 mo., and the rest in 11 mo. When can 1 pay the \\hole
witliout gain or loss ?
4. I owe $960, payable as follows : $180 in 4 mo. 20 da.,
$348 in 6 mo. 15 da., $234 in 8 mo. 5 da., and the rest in 10
mo. 13 da. Required the equated time of payment
5. A trader bought $1800 worth of goods, agreeing to pay
J of the money down, ^ of it in 5 mo., ^ of it in 6 mo., ^ of
it in 9 mo., and the rest in 12 mo. At what time may the
whole be paid ?
G. Bought a lot of goods, for which I agreed to pay
$437.75 in 3 mo., $394.25 in 6 mo., and $628.19 in 8 mo.
When may the whole be paid without gain or loss ?
7. A owes B $800, payable in 10 mo. ; but to accommo-
date B, he pays $250 down. When ought the remainder to
be paid ?
Solution. — After paying $250, he will owe $800 — $250 = $550,
wh^ch he ought to keep till its interest shall equal the interest of $800
for 10 months. Bat the interest of $800 for 10 mo., equals the interest
of one dollar for 800 times 10 mo., or 8000 mo. equals the interest of
$550 for siu of 8000 mo., which is 14^ mo. Hence it ought to be
paid in 14-^ mo.
8. I owe $1000, payable in 9 mo. ; but to accommodate
my creditor, I pay $300 down, and agree to pay $300 more in
2 mo. How long ought I, in justice, to keep the remainder ?
9. I owe $600, payable in 8 mo. 15 da., and $400, payable
in 12 mo. ; but afterwards agree to pay $400 down, and $800
in 2 mo. 20 da., on condition that I may keep the remainder
enough longer to compensato for my loss. When will the
remainder become due ?
10. A owes B $480, due in 1 yr., and B owes A $720,
due in 1 yr. 6 mo. If A should pay his debt at once, when
ought B to pay his ?
193. To find Date of Equated Time.
(a.) The best method of solving such examples as the fol-
lowing is to see how much interest will be gained or loat by
paying the sum of the debts at any assumed time.
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988 TO FIND DJLT& OF EQUATED TIHB.
(6.) It will be well as a general thing, to select for the assaoied rim«
A date on which one of the debts becomes due, as by that means we
•hall afoid the necessity of reckoning interest on that debt. Reference
■honld also be had to the probable equated time.
(tf.) The time is reckoned by counting the days between
the dates considered, as in the English method of computing
interest.
1. James Brown owes William Greene the following debts,
Ti*. : $534.88, due Jan. 7, 1855 ; $285, due April 4, 1855 ;
$327.88, due July 8, 1855 ; and $438,75, due Aug. 17, 1855.
When may the whole be paid without gain or loss ?
Solution, — Suppose that April 4, 1855, be selected as the assnmed
time. Then Mr. Brown would gain interest on
$534.83 from Jan. 7 to April 4, 88 da. » $7.84
$285.00 due at assnmed time, $0i)0
and lose int. on
$327.38 from April 4 to July 3, 90 da. = $4.91
$438.75 from April 4 to Aug. 17, 135 da. s= $9.87
S'un ^ I • $1585.96 Sum of losses = "" $14.78
debts )
Excess of losses over gains $6.94
Showing that Mr. Brown is entitled to keep $1585.96, the entire debt
due, as manj days after April 4 as it will take it to gain $6.94 inter-
est This, found by 191f is 26 days, plus a fraction less than j*.
Therefore the equated time is 26 days after April 4, which ia
April 30.
KoTB. — The above shows that on April 4th Mr. Brown could justly
haye settled the account by paying $1585.96 — $6.94 = $1579.02.
Again. (Suppose that July 3 be selected as the assumed time. Then
Mr. Brown would gain interest on
$534.83 from Jan. 7 to July 3, 178 da., = $15.86
$285.00 from April 4 to July 3, 90 da., = 4.27
$327.38 due at assumed time, 00.00
Giving for sum of gains, $20.13
and lose int. on
$438.75 from July 3 to Aug. 17, 45 da., = 3.2»
Sum of I
debta. J *** $1585.96 Excess of gain over loss, $16.84
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TO FIND DATE OF EQUATED TIME. 289
Showing that Mr. Brown ought to pay $1585.96, the entire deht dne, as
many days before July 3 as it will take it to gain $16.84 interest This,
found as before, is 64 days nearly. Therefore the equated time is 64
days before July 3, which is April 30, as before.
NoTB. — The aboye shows that if the account should not be settled
till July 3, Mr. Brown ought justly to pay $1585.96 + $16.84 a
$1602.80.
Proof. — By paying the debt on April 30, Mr. Brown will gain in
tar3St09
$534.83 from Jan. 7 to April 30, 114 da.=: $10.16
$285.00 from April 4 to April 30, 26 da. = 1.23
Making sum of gains = $11.39
He will lose interest on
$327.38 from April 30 to July 3, 64 da. = $3.49
$438.75 from April 30 to Aug. 17, 109da.= $7.97
Making sum of losses, = $11.46
Excess of loss over gain, = $00.07
^ wbich, being less than the interest of $1585.96 for a half day, shows
2ihat April 30 is the correct equated time.
^ 2. I owe $387.53, due Nov. 7, 1851 ; $467.81, due Dec
21, 1851 ; $256.19, due Feb. 11, 1852 ; $136.43, due March
1, 1852; and $387.59, due May 3, 1852. What is the
cqt»ted time of payment ?
3. I owe $2867, due April 15, 1850 ; $1642, due July 27,
1850 ; $4371, due Oct. 8, 1850 ; and $5940, due Jan. J,
1851. What is the equated time of payment ?
4. I owe $628.13, due Dec. 17, 1852 ; $427.19, due Dec.
23, 1852 ; $371.16, due Dec. 30, 1852 ; $587.83, due Jan. 3,
1853 ; $987.62, due Jan. 7, 1853 ; and $843.28, due Jan. 14,
1853. What is the equated time of payment ?
How much is due on the above Jan. 1, 1853 ?
5. I owe $543.28, due April 24, 1855 ; $723.13, due May -
11; 1855 ; $484, due Sept. 3, 1855 ; $426.18, due Oct 10,
1855 ; $236, due Nov. 10, 1855. What is due on the above
Sept. 1, 1855, interest being reckoned at 5 per cent? -^ ^T
6. What is the equated time for paying the following
debts : $600, due March 7, 1850 ; $400, due June 11, 1850 1
25
^
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t90 EQUATION OF ACCOUNTS.
$800, dao Aug. 17, 1850; $500, dae Oct 8, 18^0; acd
$1000, due Nov. 27, 1850 ?
194I* JSquaUon of Jeeounts. v
(a.) The method of finding the equated time w^en each
jMuiy owes the other, that is, when there are entries on boih
the debit and credit side of an accoimt, does not difier in prin-
ciple from that in which there are entries only on one side.
Tlie following example and solution will illustrate it : —
1. The account books of A and B show that /w
.A^ CSV-
$426.70, due Jan. 6, 1855.
$41S.65, due Feb. 2, 1855.
$169^8, due April 13, 1855.
$328.57, due Aug. 29, 1855.
jlHd ilini IT •mm i
$148^7, due Dec. 22, 1854.
$173.19, due Jan. 29, 1855.
$587,23, dne May 7, 1855.
$658.45, due Sept 30, 1855.
When ought the balance to be paid ?
S[)/ufum. — Suppose that April 13, 1855, be the assumed time of pay>
j:ient Then A will gain interest on each of bis debts which becomes
dne to B before that time, and on each of B*s debts which become due
Co him after that time ; for he will hare the use of each for a longer
time than he is justly entitled to. He will lose interest on each of his
debts which becomes due to B after that time, and on each of B*s debts
which becomes due to him before that time ; for he will not have the
use of them for so long a time as he is justly entitled to. Hence A will
gain the interest of
$426.70 from Jan. 6 to April 13, 97 da., » $ 6.90
$413.65 from Feb. 2 to April 13, 70 da., == $ 4.83
$169.28 from Feb. 13, $0.00
$587.23 from April 13 to May 7, 24 da., » $ 2.35
$658.45 from April 13 to Sept 30, 170 da., <es $18.65
SumofgrmsB . . . $32 7S
A will lose the interest of
$328.57 from April 13 to Aug. 29, 138 da., = $7.56
$148.37 from Dec. 22, 1854, to April 13, 1855, 112 da., = $2.76
$173.19 from Jan. 29, to April 13, 74 da^ » $2.14
iSum of losses, $12.46
Excess of A's gain over his loss, or of B's loss oyer his gain, $20.27
y Google
Digitized by*
Situation of accounts. 291
But the sum of A*s debts is $1338.20, and of B*8 is $1567^4.
$1567.24 — $1338.20 = $229.04, the balance which B owes A.
The question now resolves itself into this : If by B's paying A
$229.04 April 13, 1855, A gains and B loses $20.27 interest, when can
he pay ii without any gain or loss of interest 1 The answer evidently
is. As many days after April 13, 1852, as it will take $229.04, or, disre-
garding the cents, $229, to gain $20.27 interest This, found by methods
bef }ro explained, is 531 days = I yr.* 166 da., and shows the equated
time to be Sept. 26, 1853, which may be proved as were the former ex-
auples.
Note. — Although accounts like the above are sometimes settled by
notes palpable at the equated time, they are more frequently settled by
notes pa3able at some more convenient time, or by cash. In all such
cases, allowance is made for the interest gained or lost Thus, if the
above account should be settled by cash April 13, 1852, $20.27 would
be deducted from the balance due from B to A, in order to compensate
B for the interest he would lose ; that is, B would pay A $229.04 —
$20.27 = $208.77. If it should be paid May 1, 1852, B would have to
pay A $.69 (the interest of the balance due A from April 13 to May 1 )
more than if he had paid it April 13 ; or, which is the same thing, he
would have to pay the balance $229.04, minus its interest $19.58, from
May 1, 1852, to the equated time. If the balance due at any given time
had been originally required, it should have been found directly by mak-
ing the given time the " assumed time."
2. By the respective accounts of Henry Lane and William
Pond, it appears that
Pond owes Lane
$876.37, due April 5, 1852.
579.48, due May 3, 181)2.
487.83, due June 11, 1852.
145.38, due Aug. 8, 1852.
$2089.06 s= amt due Lane.
And that Lane owes Pond
$228.13, due April 28, 1852.
347.16, due June 3, 1852.
313.27, due July 28, 1852.
839.42, due Sept 1, 18.52.
$1727.98 = amt due Pond.
$2089,06 — $1727.98 = $361.08 = balance due Lane.
When can this balance be paid without gain or loss to
eitber party ?
Solution. — Suppose it to be paid June 11, 1852. Then will Bfr
Pond gain the interest of —
^ Beckoning the year as 365 days, as is always done in such cases, un*
less it ncludes February of leap year, when it is leckoned as 366 diys.
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X92
EQUATION OF ACCOUNTS.
•876^7 from April 5 to June 11, 67 da^ » $ 9.79
579.48 from Mny 3 to June II, 39 da^ *= 3.77
487.83, dae June 11, a o.OO
813.37 from Jane 1 1 to July S8, 47 da., « 2.45
83942 from June 1 1 to Sept. 1, 82 do., » 11.47
8am oi gainf ^
He will lose the interest of
$145,38 from June 1 1 to Aag. 8, 58 da., b $1^1
228,13 from April 28 to June 11, 44 da., « 1.67
847,16 from June 3 to Jane 11,8 da^ a ^6
$27.48
Sam of losses, . . $ 3.54
Excess of gain over loss, .... ... 23.94
As Mr. Fond gains this interest on money which he owes, he onght
to pay the deht ($361.08, the balance of the account) as many days be-
fore June 11, 1852, as it will Uke for it to gain $23.94 interest This
gives for the equated time 398 days before Jane 11, 1852, which is May
10, 1851. The sum necessary to settle the account after the equated
time will be the amount of the balance, $361.08, from the equated time
to the time of settlement
8* When was the balance of the following account due ?
Dr. George Ide, in account with James Snow. Cr.
1849.
1849.
—- —
■
Jan. 17.
To Mdse. .
$336
18
Feb. 1.
By Mdse. .
$421
30
Jan. 31.
To Mdse. .
443
17
Feb. 27.
By Mdse. .
620
00
March 7.
To Cash, .
218
63
Mar. 13.
By Mdse. .
283
17
April 17.
To Mdse. .
500
00
April 29.
By Mdse. .
482
29
May 28.
To Mdse. .
84
36
June 1.
By Mdse. .
825
13
4. When was the balance of the following account due ?
Dr. George Black in account with John Brown. Cr.
1850.
18,50.
May 13.
To Mdse. 4 mo.
$431
17
June 1.
By Mdse. 3 mo.
$223 62
^•ly 25.
To Mdse. 3 mo.
256
38
July 7.
By Mdse. 6 mo.
150 00
^ Anp. 8.
To Mdse. 6 mo.
431
72
July 22.
By Mdse. 3 mo.
250J00
Bcpt 23.
To Mdse. 3 mo.
585
41
Sept 1.
By Cash,
300 ; 00
Hk^Y. 7.
To Mdse. 3 mo.
738 16
Nov. 23.
By Mdse. 2 mo.
138^16
Dec. 1.
By Mdse. 3 mo.
182i81
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TO PIND\ THE PRINCIPAL, OR INTEREST.
293
Note. — 4 mo., 3 mo., &c., means that goods were sold at so many
months* credit.
C-. What was due on the following account Jan. 1, 1853 ?
Dr. George Mann in account with Henry Guild. Cr.
1S52.
I
1852.
—
May 5.
To Bal. 3 mo.
$513|43
April 20.
By Mdse. 6 mo.
S328
13
Juno 27.
To Mdse. 4 mo.
624
27
May 10.
By Mdse. 3 mo.
143
27
July 3.
To Mdse. 6 mo.
831
13
June 13.
By Mdse. 4 mo.
837
19
Bcpt. 5.
To Mdse. 2 mo.
47
62
July 7.
By Mdse. 6 mo.
56
18
Sept 17.
To Mdse. 3 mo.
125
53
Aug. 20.
By Mdse. 4 mo.
123
42
Oct. 19
To Cash,
387
00
Oct. 1.
By Mdse. 3 mo.
78.30
Dec. 1. To Cash,
629 28 1
Nov. 23.
By Mdse. 2 mo.
127|14
6. What was due on the following account Jan. 1, 1853,
interest being 7 per cent, and 4 months' credit being allowed
on each entry ?
Dr. David H. Daniels in account with George W. Deane Cr.
1852.
i
1852.
—
July 8.
To Mdse. .
S 236
^'i
July 3.
By Mdse. .
$439
27
Aug. 1.
To Sundries,
819
63
July 25.
By Mdse. .
213
16
Sept. 4.
To Mdse. .
142
13
Sept. 13.
By Mdse. .
100
00
Nov. 13.
To Mdse. .
947
22
Oct. 24.
By Mdse. .
262
18
Dec 8.
To Sundries,
1050 OOj
Nov. 30.
By Mdse. .
327
48
1
Dec. 21.
By Mdse. .
520
75
19«S« To find the Principal, or Interest, from the Amount^
Hate, and Time.
(a,) When the amount, time, and rate are given to find the
principal or interest, we find what part any principal, or (if
tlie interest be required) its interest for the given time, at the
given rate, is of its amount, and then take this part of the
pven amount.
(5.) The first step towards this is to find the fraction ex-
pressing what part any interest for the given time, at th«
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294 TO mCD THE PRINCirALy OB I1ITEBB9T.
given rate, h of its principal. This fraction will aiwatf:^ be
the some part of the given annual rate that the given lime is
of 1 year, or 360 days ; or, if the rate is 6 per cent, it wiB
eqnal t) e fraction expressing the part which the given time is
of 200 montlis, or 6000 days. The amount, of course, will
equal the principal, plus the fractional part of it which the in*
terest equals.
(c.) Thus, interest for I yr. 7 mo^ or 19 months, at 6 per cent pet
yoAT, as 3^ of the principal, and the amount ss J^ -f. ^^^ = ^^
of the principal Hence, 7^ of the principal = 7^ of the amonnt,
and the entire principal =: J^, and the interest 3^, of &e amount
lOr 19 mo. at 6 per cent (The same fractions would hare been obtained
by considering the interest to be ^-f of t^ of the principal.)
(rf.) Again. Interest for 19 months at 4^ per cent per year == ^
of jQQ s=s if of 7^ s= if^ of the principal, and the amount = %%%
+ 1% = f W of the principal. Hence, the principal = f^, and the
interest =? ^ft^, of the amount for 19 mo. at 4^ per cent
(e.) Again. Interest for 2 yr. 3 mo. 2 da., or 812 days, at 6 per cent
per year, = ^j\nr = i^pS of the principal, and the amount = \% %Z
+ T^Vu" = \i%i of the principal. Hence, the principal = -ff J^, and
the interest = -^^W* of the amount for 2 yr. 3 mo. 2 da. at 6 per cent
(The same fractions would have been obtained by considering the inter-
est to be 5^8 of T^TT of the principal.)
(/) Again. Interest for 5 mo. 14 da., or J64 days, at 7 per cent per
year, = ^^ of ihs = fijr of yirr = iS^ of the principal, and the
amount = %%%% + ^^^ = fffj of the prindpaL Hence, the prin
cipal = I^S^, and the interest = ^^, of the amount for 5 mo. 14
da. at 7 per cent
1. What principal on interest at 6 per cent per year wiH
amount to $884,125 in 1 yr. 2 mo. 10 da. ?
Solution. — Since, at 6 per cent per year, interest for 1 day «= ^lAny
of the principal, interest for 1 yr. 2 mo. 10 da., or 430 days, must equal
FcAftr, or ^^, of the principal, and the amount must equal fJJ +
w(nTi or fro* of the prir.cipal. Hence, -^^ of the principal must equal
^St and the principal itself mnst ecjual fj^ of the amount fj^- 0/
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TO FIND THE PRINCIPAL, OR INTEREST. 295
'$884,125 = $825 = principal required. The mere numerical work
may be indicated thus : —
1 yr. 2 mo. 10 da. = 430 da.
^Sj%% or M.+m=-ui
f £S of $884,125 = $825 = principal required.
Second Solution. — The amount of $1 for 1 yr. 2 mo. 10 da. = $1.07^,
and if one d-Uar is required to gain $1.07^, as many dollars will be
required to g^ain $884,125 as there are times $1.07iJ^ in S884.125 But
6884.125 -^ $1.0'?^ = $884,125 -J- f H = If 7 of $§all25 = $825, as
before. \
Proof. — The amount of $§25 for 1 yr. 2 mo/lO da. is $884,125
2. What principal on intd^est at Z^per cent per year will
amount to $703,551 in 4 mo. ^7 da,^'^
Solution. — At 7 per cent per ycfir^ interest for 4 mo. 27 da., or 147
days, = -J^ of tJtj- == T^Mxy of "ae principal, and the amount =a
if S8S + Tf^^TT == if a^S of /the Vrincipal. Hence the principal
must equal if ^f S o{ the amount, whU;h in this case is if ^J of
$703,551 = $684. The mere numerical vrhrk may be indicated thus : —
4 mo. 27 da. = 147 da* ^H of tStj V lihiv-
HHi of $703.5^1 = $684 = principal Wired.
Second Solution, -r- The amount of $1 at 7 peX cent for 4 mo. 27 da.
8=$1.02i3^; and if one dollar is required to gainV$l.02i^^, as many
dollars would be required to gain $703,551 as there are. times $1,02^^ in
$703,551. But $703,551 -5- $l.02iJ5 = $703,551 -5- ifJM == \l%i%
of $703,551 = $684, as before.
"Proof. — The amount of $684 for 4 mo. 27 da. at 7 per cent equali
1703.551.
"What principal will amount to —
8. ^ $569,296 in 8 mo. 16 da. at 6 per cent?
4. $573.16 in 2 yr. 9 mo. 10 da. at 6 per cent ?
5. - $922.13 in 1 yr. 4 mo. at 8 per cent ?
e. , $378.82 in 1 yr. 4 mo. 20 da. at 6 per oent ?
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296 DISCOUNT AND PRESENT WORTH.
7. $57.72 in 8 ma 20 da, at 10 per cent ?
8. $899,944 in 5 mo. 14 da. at G per cent ?
198. Discount and Present Worth.
(a.) Discount, as it is technically called, furnishes the
most common application of the processes of the preceiing
article to the problems of business life.
(b.) It is obvious that the true present value of a debt due
at a future time is that sum of money which, put on interest
at the present tune, will amount to the given debt at the time
it becomes due.
Thus, when the rate of interest is 6 per cent per rear, a debt of $106
doe in one year is worth the same as a debt of $100 due now ; for if the
money received on the second debt be put on interest, it will amount to
$106 in one year ; that is, when the first debt becomes duo.
(c.) A debt due at a future time may be regarded, then, as
the amount of a principal on interest from the present time
to the time when the debt will become due. This principal
is usually called the present worth of the debt, and its
interest is called the discount, because, if discounted or
deducted from the debt, it leaves the present worth.
(d.) From this, it follows that to ask what is the present worth of
$651, due in 6 mo. 20 da., money being worth 6 per cent per year,
Is equivalent to asking what principal on interest at 6 per cent will
amount to $651 in 6 mo. 20 da. ; and that the solution of the first qnes-
tion is the same as that of the second. To test the correctness of any
resnlt, see if the amount of the present worth equals the given debt.
1. What is the present worth of $438.18 due in 1 yr. 6
mo. at 6 per cent per year ?
Partial Solution. — The present worth reqnired is that sum of money
which, put on interest at 6 per cent, will amount to $438.18 in 1 yr. $
m.1., or 18 mo. The interest for 18 mo. = ^^^ly = xS^y of the prind
pal, and the amount =, &c., as m the last article.
What is the present worth of —
2, $83.45 due in 4 yr. 2 mo. at 6 per cent ?
8. $89.88 due in l~y \ at 7 per cent?
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DISCOUNT AND PBESENT WORTH. 297
4 $142.56 due 2 yr. hence at 4 per cent ?
5. $122.94 due 4 mo. 27 da, hence at 6 per cent ?
"^6. $475.64 due 1 30*. 8 mo. hence at 6 per cent ?
7. $578.50 due 3 yr. 1^ mo. hence at 8 per cent ?
8. $731.52 due 3 yr. 4 mo. hence at 6 per cent ?
9. $1823.70 due 7 mo. 15 da. hence at 5^ per cent?
iO. What is the discount of $195.87 due 1 yr. 5 mo. 19
dx* hence, at 6 per cent per year ?
Direction. — Find what part of the debt the discount is, and get that
part of $195.87. For proof, subtract the discount thus found from
$195.87, and see if the interest of the remainder for the given timo
equals the discount. We may also get the discount by finding the
present worth, find subtracting it from the debt ,
What is the discount of —
-.11. $394(5.11 due 2 yr. 5 mo. 15 da. hence at 6 per cent?
12. $6392.43 due 15 mo. 7 da. hence at 6 per cent ?
13. $1241.27 due 1 yr. 5 mo. 23 da. hence at 6 per cent?
14. $6255 due 3 yr. 2 mo. hence at 5 per cent?
^5. $179.96 due 2 yr. 3 mo. 6 da. hence at 4 per cent?
16. I own a note for $97^6, payable on demand with in-
terest, and another for $1034.56, payable in just 1 year, with
interest afterward. Allowing money to be worth 6 per cent
per year, ^hich debt is justly worth the most at the present
time, and how much the most? Which will be worth the
most at the end of the year ? -^hich will be worth the most
at the end of 6 months ? Which will be worth the iiiost at
the end of 2 years ?
17. If two notes are given on the same day, one for a cer-
tain sum due at a future time, with interest afterwards, and
the other for the true present worth of the first note, payable
on demand with interest, their true values will be the same at
the time they are given, and also at the time the first be-
comes due ; but at all other times they will differ. At any
time between the day of their date and the day when the first
note becomes due, the true value of the second note will be
greoiPK thait that of the first ; but at any time after the first
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298 BUSINKSU JCKTHOU OF DISCOOKT.
note becomes due, the true value of the first note will bi
greater tlom that of the second. Show why this is so.
19T* Business Method of Discount
(a.) Business men are usually willing to allow on money
paid for goods before it is due, a discount equal to, or greater
than, its interest from the time of payment to the time when,
by the conditions of the sale, it would have become due.
Thus, when only the interest of the debt is disconnted. S824 dnft in
6 months, interest being 6 pft cent per year, is regarded as worth $824
— 03 of $824 =s $824 — $24.72 = $799.28, whereas it ought to be worth
$800.
(6.) This is always an advantage to the person owing the money, as
it enables him to pay his debt for less than the sum which, pnt at inter
est, would amount to the debt at the time it would become due.
(c.) It is common to deduct as much as five per cent from the face
of a bill due in four or six months, and evcu more is sometimes de*
ducted.
(cL) The present worth thus obtained may be called the
ESTIMATED PRESENT WORTH, tO distinguish it from the TRtTE
PRESENT WORTH, obtained by the method explained m 196.
1. I owed a debt of $8692, payable June 1, 1852 ; but my
creditor, offering to allow me discount estimated by the busi-
ness method at the rate of 6 per cent per year, if I would
pay the debt Jan. 1, 1852, 1 borrowed money for the purpose
at G per cent interest. June 1, 1852, 1 paid the amount of
the borrowed money. What was my gain by the transac-
tion'^?
2. Owing a debt of $1545, due in 6 months, when money
h worth 6 per cent per year, what shall I gain by hiring
money enough to pay it now, allowing the usual business dis-
count on the debt, and then paying the borrowed money with
interest, when the original debt would otherwise have become
due?
"^ 3. At 6 per cent per year, what is the difference between
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BUSINESS METHOD OP DISCOUNT. 299
the bank discount and the true discount of a note for $2059.40,
payable in 60 days ? ♦
4. Keceived for my note of S600, payable in 6 months, its
true present worth. How much more did I receive on it than
I should have received at a bank, money being worth 6 per
cent ? How much interest money shall I have gained, when
the note becomes due, over what I should have gained on th'e
present worth, as determined at the bank ?
'^ 5. For how much must a note, payable in 30 days, be
given, that, when discounted at a bank, $900 may be received
on it, money being 6 per cent ?
Solution. — The money received on a note dbccmnted at a bank
equals the sum for which the note is given, minus its interest for the
time before it becomes due. Since, at 6 per cent, the interest for 30
days and grace, or 33 days, = ^S^ = -ziitu of the principal, the sum
received must equal m% — 'siixf = iSSS" of the face of the note.
Therefore the face of the note = HH of the sum received on it,
^= Hi% of $900 =s $904,977, or, as it would in practice be considered,
$904.98.
The numerical work can be expressed thus : —
83 Jl_ 2000 __ _11_ ^ 1989 2000 _^ 200000 ^
6000 '^ 2000' 2000 2000 2000' 1989 ® '^ 221 "*
$904.977 ;
kence &ce of note := $904.98.
6. How much would be received at a bank on a note for
$904.98, payable in 30 days ?
KoTB. — The above example suggests the method of proving the 5th.
7. For how much must a note payable in 3 months be
l^vcn, that, when discounted at a bank, $1000 may be re-
ceived on it, money being worth 6 per cent per year ?
8. For how much must a note payable in 6 mo. bv> ^ven,
that, when discounted at a bank, $1800 may be received on
it, money being worth 6 per cent per year ?
* Eemomber that three days' grace are always allowed on a note, unUss
the contrary is specified.
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aOO TO FIND TnE BATS.
9. Obtained at a bank, on my note payable in 6 mo., mooej
enoagh to buy 20 acres of land at $100 per acre. The day
my note at the bank became doe, I sold the land for $2062.^2
cash. Did I gain or lose by the transaction, and how much,
money being worth 6 per cent per year ?
10. Obtained at a bank, on my note payable in 4 months,
money enough to buy 20 acres of land at $100 per acre.
The day the note became due, I sold the land for cash, at such
rate that the price of 18 acres was just sufficient to pay the
note. How much did I gain by the transaction, money being
worth 6 per cent ?
11. Bought goods to the amount of $86427 on a credit of
6 months ; but the seller offering to deduct 5 per cent from
the face of the bill if I would pay cash, I hired the requisite
amount of money, giving my note payable in 6 months, with
interest at 6 per cent per year, to be reckoned from date.
For how much less than the value of the original bill could I
pay the amount of this note ?
12. I owed $800, due in 6 months ; but my creditor offer-
ing to deduct 5 per cent of the debt for cash, I paid $380
down. How much did I still owe ?
Suggestion. — Since 5 per cent of the debt was to be dedacted for
cash, the cash payment wonld be 95 per cent, = -fi^ «=» ^, oT the part
of debt it would cancel ; or the part cancelled would be ^ of the cash
paid.
13. I owed $900, payable in 4 months ; but my creditor
offering to deduct 4 per cent of the debt for ready money, I
paid $696 down. How much did I stiU owe ^
198. To find the Srtie.
Problems in which, the principal, interest, and time being
given, we are required to find the rate, rarely occur in busi-
ness life. The following solutions illustrate the prindplei
which apply to them.
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TO FIND THE PRINCIPAL. 801
1. At what rate per cent must $648 be on inte/est to gain
4^1.873 in 2 yr. 3 mo. 17 da. ?
Solution, — The principal being eqnal to 648,000 mills, and the inter-
est tc 81,873 mills, the interest is ^jVVrpTT = ^lyJu of the principaL
8 yr. 3 mo. 17 da. == 27 mo. 17 da. = 827 da. If the interest for 827
da. = T^^V of the principal, the interest for 1 day must equal f^Jy
of ^y^jj of the p»nncipal, and the interest for I yr., or 360 da., must
equal 360 times the last result, or f J^ of T^^^yVxy = sWr — -05^ =» 5^
per cent.
Required the rate of interest when —
2. $624 gains $74.88 in 1 yr. 2 mo. 12 da.
3. $57.25 gains $5,038 in 1 yr. 5 mo. 18 da.
4. $855 gains $46.55 in 2 yr. 2 mo. 4 da.
5. $64.80 gains $6,246 in 11 mo. 17 da.
199. To find the Principal from the Interest.
* Problems in which, the interest, rate, and time being giTcn,
we are required to find the principal, are, like those in the
last article, of rare occurrence.
1. What principal on interest at 6 per cent will gain
$37.47 in 1 yr. 3 mo. ?
Solution, — At 6 per cent per year, the interest of any principal for
15 mo. s= yVcF = T^ of the principal. If $37.47 is ^ of the princi-
pal, ^ of the principal must equal i of $37.47, and the principal must
equal 40 times the last result, or ^ of $37.47, which is $449.60.
2. What principal on interest at 8 per cent wiU gain
$26.18 in 1 yr. 4 mo. 15 da. ?
Solution. — Since 1 yr. 4 mo. 15 da. = 16^ mo. = ^^ yr. = -V^ of
1 yr., the interest must equal JLL of 8 per cent, or 11 per cent of the
principal. If $26.18 is 11 per cent of the principal, 1 per cent of tbo
principal must be i^ of $26.18, which is $2.38, and 100 per cent, or the
principal, must equal 100 times the last result, which is $238.
What principal on interest —
8. At 6 per cent will gain $8.73 in 5 mo. ?
4.' At 6 per cent will gain $4.77 in 1 yr. 5 mo. 20 da^ ?
26
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8f/S COMI*OUf«l> INTEREST.
5. At 5 per cent will gain $ 1.27 in 2 yr. 6 mo. ?
6. At 7^ per cent will gain $116,127 in 4 yr. 4 mo. 4, cla-f
300 • Compound Interest,
When interest is to be paid at regular intervals, or, if iin-
paid, is to be added to the principal, to form a new principal
on which interest is to be computed, it is called Compound
Interest. The following example illustrates this : —
1. What is the compound interest of $784 for 2 jr. 8 mow
At 6 per cent, payable annually ?
Solution,
a = $784. = principal.
06 of a = b =s 47.04 s= interest for Ist year.
a + b!=c= 831.04 = amonnt due at end of 1st year.
X)6 of c = d = 49.862 = interest for 2d year.
c + d = e = 880.902 = amount due at end of 2d year.
^ of e s= f =3 35.236 = interest for 8 mo.
6 + f = g = 916.138 = amount due at end of 2 yr. 8 mo.
a =s 784. = principal.
g — a = h =s $132,138 = compound interest for 2 yr. 8 mo.
2. To what sum will $437 amount in 2 yr. 6 mo. at 4 par
fent, payable semiannually ?
Solution:
a s= $437 = principal.
jOS of a = b := 8.74 = interest 1st 6 mo.
a + b s= c = 445.74 = amount at end of 6 mo.
i)2 of c = d =s 8.915 s= interest 2d 6 mo.
c -f- d = e = 454.655 = amount at end of 12 ma
X)2 of e = f == 9.093 = interest 3d 6 mo.
e + f = g = 463.748 = amount at end of 18 mo.
02 of g = h = 9.275 = interest 4th 6 mo.
i -I- h = ! = 473.023 = amount at end of 2 yr
^ ^ i = j ^ Q.iff '^ JntercKt 5th 6 mo.
i- j -» k *» *.1?{?.483 = «ujuva-i. crue at and of S vr f :
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TABLE FOR COMPOUND INTEREST.
803
3. What is the compound interest of $938.63 for 4 yr. 6
mo. at 6 pel cent, payable annually ?
4. What is the compound interest of $573.32 for 2 yr. 3
xno. at 8 per cent, payable quarterly ?
5. To what sum will $1000 amount in 3 yr. 2 mo. at 6
per cent, payable annually ?
6. To what sum will $500 amount in 4 yr. 3 mo. at 5 per
cent, payable semiannually ?
301* Table for Compound Interest
Tahh showing what part of any principal is equal to it$
amount at 3, 4, 5, 6, 7, and 8 per cent annual compound
interest, for any number of years not exceeding 25.
Yre.
1.
3 per cent.
4 per cent.
5 per cent.
6 per cent.
7 per cent.
8 per cent.
1.03
1.04
1.05
1.06
1.07
1.08
2.
1.0609
1.0816
1.1025
1.1236
1.1449
1.1664
3.
1.092727
1.124864
1.157625
1.191016
1.225043
1.259712
4.
1.125509
1.169858
1.215506
1.262477
1.310796
1.360489
5.
1.159274
1.216653
1.276281
1.338225
1.402552
1.469328
6.
1.194052
1.265319
1.340096
1.418519
1.500730
1.586874
7.
1.229874
1.315932
1.407100
1.503630
1.605781
1.713824
8.
1.266770
1.368569
1.477455
1.593848
1.718186
1.850930
9.
1.304773
1.423312
1.551328
1.689479
1.838459
1.999005
10.
1.343916
1.480244
1.628894
1.790848
1.967151
2.158925
11.
1.384234
1.539454
1.710339
1.898298
2.104852
2.331639
12.
1.425761
1.601032
1.795856
2.012196
2.252191
2.518170
13.
1.468534
1.665073
1.885649
2.132928
2.409845
2.719624
14.
1.512590
1.731676
1.979931
2.260903
2.578534
2.937194
15.
1.557967
1.800943
2.078928
2.396558
2.759031
3.172169
16.
1.604706
1.872981
2.182874
2.540351
2.952164
3.425943
17.
1.652848
1.947900
2.292018
2.692772
3.158815
3.700018
18.
1.702433
2.025816
2.406619
2.854339
3.37993.
3.996019
19.
1.753506
2.106849
2.526950
3.025599
3.616527
4.315701
20.
1.806111
2.191123
2.653298
3.207135
3.869684
4.660957
21.
1.860294
2.278768
2.785962
3.399563
4.140562
5.033834
a2.
1.916103
2.369918
2.925260
3.603537
4.430402
5.436540
23.
1.973586
2.464715
3.071524
3.819749
4.740530
5.871464
24.
2.032794
2.563304
3.225100
4.048934
5.072367
6.341181
25.
2.093778
2.665836
3.386355
4.291870
5.427433
6.848475
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t04 COMMISSION.
Use of the Table. — The amount of any sum of monej,
at compound annual interest, for any time and rate mentioned
in the table, may be found by multiplying the principal by the
appropnate number selected from the table. This will be
illustrated in the following examples and solution.
1. What is the amount of $245.73 for 12 yr. 6 mo. 20 da.
annual compound interest at 6 per cent ?
SoiutUm, — Bj the table, it appears that the amount of a sam fog
12 years, nt 6 per cent, cotnpcmd interest, b 2.01 21 9G times the princi-
4^1. Multiplying $245.73 by 2.012196, we have $494.45692308, or, omit-
ting the denominations below mills,
a es $494,457 3= amount for 12 years.
^ of a Ks b ea 16.481 s= interest for 6 mo. 20 da.
A ^ b OB c Bs $510,938 = amount for 12 yr. 6 mo. 20 da
What is the amount at annual compound interest of -^
2. $578.67 for 11 yr. 4 mo. at 3 per cent ?
8. $147.43 for 22 yr. 4 mo. 24 da. at 5 per cent'?
4. $1467 for 18 yr. at 7 per cent ?
What is the compound interest of —
5. $1 137.88 for 13 yr. 6 mo. at 6 per cent ?
6. $828.96 for 9 yr. 3 mo. at 4 per cent ?
303. Commission.
Money received for services in buying and selling goods
for others is called Commission, and is usually reckoned at a
certain per cent of the cost of the goods bought, or price of
hose sold. A merchant who makes it his business to buy
and sell on commission is called a commission merchant.
(See 177th page, Ex. 21, Note.)
1. A commission merchant sold goods for a manufactnrer
to the amount of $568.36, for which he charged a commission
of 2J. per cent. What was his commission, and how much
will be due to the manufacturer ?
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COMMISSION. 305
Answet' r— His commission = .02 J of" $568.36 = $M.209, and the
gam due the manufacturer = $568.36 — $14.21 = $554.15.
2. A commission merchant bought goods for a country
trader to the amount of $738.27, for which he charged a com-
mission of 1^ per cent What was his commission, and what
sum must the trader remit to pay for the goods and commis-
sion?
Answer. — His commission = .01^ of $738.27 == $11.07, and the
trader must remit $738.27 + $11.07 = $749.34.
8. I sold for Seth Jones 2024 pounds of butter at 19 cents
per pound, and 5276 pounds of cheese at 7^ cents per pound.
What was the vahie of ray commission at 2^ per cent on the
Eales ? How much money owght 1 to pay him ?
4. I bought for Francis Jackson 50 pairs of boots at
$3.87^ per pair, 100 pairs at $2.75 per pair, 75 pairs at
$2.J6f per pair, 100 pairs of shoes at $1.56 per pair, and 80
pairs at $1.08 per pair. What was the value of my commis-
sion at 2 per cent on the purchase ? How much money must
Mr. Jackson remit to me to pay for the goods and my com-
mission ?
5. 1 have sent $5000 to my agent in New Orleans, direct-
ing him to expend it for cotton, first deducting his commission
of 2 per cent on the purchase. What will be his commission,
and what will he expend for cotton ?
Suggestion. — The $5000 sent includes the sum to be actually invested
and my agent's commission of 2 per cent on that sum, and is therefore
•fj§ of the purchase money. Hence, the purchase money, or sum to be
expended by my agent, is ^%^ of $5000, and his commission is xSj of
$5000. For proof, see if the sum expended, plus .02 of it, equals $5000.
6. I have received $5600 from my correspondent in St
Louis, with directions to expend it in merchandise, first de-
ducting my commission of 2^ per cent on the money expend-
ed. How much ought I to expend ?
7. My agent in Rochester writes that he has purchased a
lot of flour for me at $5 per barrel, and that the entire cost
26*
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806 coxMiasioN.
of the flour, including bis commission of 2^ per cent, is $1 600
How many barrels of flour were purcbaeed ?
8. Briprjrs Sc Grant sold for Jenks, Clarke, & Co. 1000
brooms at S.25 apiece, for whicb they charge a commission of
4 per cent. Pursuant to instructions, they invest the balance
in sugar at 8 cents per pound, first deducting their commission
of 2 per cent on the purchase. How many pounds of sugar
did they buy ?
9. My agent at Cincinnati writes that he has purchased on
my account a lot of provisions, and that his commission of 1 j^
per cent on the purchase is $13.50. How many doUam
worth of provisions has he bought ?
10. My agent at New Orleans has purchased for me a lot
of cotton at C cents per pound, for which he charges a com-
mission of 1 J per cent. His commission amounts to $38.80.
How many pounds of cotton has he bought, and how much
money must I remit to him to pay for the cotton and his com-
mission?
11. Haskell & Latham sell for me, at auction, goods to the
amount of $8732. Their charges are as follows : Auction
tax, 1 per cent; commission, 4 per cent; guarantying the
sales, 3 per cent ; advertising, $23.25 ; truckage and storage,
$1 1.25. How much money will be due me ?
Suggestion. — The auction tax, commission, and gnarantv amount to
8 per cent of the value of the goods, which, together with the other ex-
penses, must be deducted from the value of the goods, to leave the sum
due me.
12. Lewis & Johnson sell for Field & Dean 96 cases of
cassimere, each case containing 276 yards, at $1.25 per yard,
for which they charge a commission of 3^ per cent. They
receive instructions from Field & Dean to remit them half of
the net proceeds, and to invest the remainder, after deducting
a commission of 1^ per cent on the purchape, in wool, at 50
cents per pound. What was the value of the cash remitted
♦o Field 8c Dean ? How many pounds of wool were bojight?
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STOCKS. MT
303. Stocks.
(tt.) Money invested in any property designed to yield an
Y tticome is called stock.
Tho money invested by a man in his business is called his stock nr
tjiadb; that invested in government securities, bonds, &c^ is called
OOYE&NMBNT STOCK.
(ft.) The CAPITAL STOCK of any incorporated company is
tlie money paid in by its members for the general purposes
for which the company was formed. It is divided into equal
parts, called shares. Any person owning one or more of
these shares is a stockholder, or member of the corpora-
tion. Stocks is a general term applied to the shares them-
selves, which may be bought or sold like any other property.
(c.) The value at which the shares of any corporation are
rated in estimating its capital stock, that is, their first or origi-
nal value, is called their nominal value, or their par value,
and is always the same. The price which they will bring, if
exposed for sale, is their true or real value, and is differ-
ent at different times. If the real value equals the par value,
the stocks are at par ; if it be greater, they are above par,
and sell at a premium, or advance ; if it be less, they are
below par, and sell at a discount.
(c?.) The profits accruing to the corporation, if any, are at
intervals distributed among the members in proportion to the
number of shares each holds, and are then called dividends.
The dividends are usually reckoned at a certain per cent of
the par value of the shares.
1. How much will 11 shares Fall River Haihroad stock
cost, at an advance of 6 per cent, the par value being $1(^0
per share ?
Solution,
$1100 = par value of II shares.
66 = 6 per cent premium.
$1166 = real value, or required cost.
2* How mu'^h will 11 shares Providence Railroad stoob
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M6 STOCKS.
ensty at a discount of 6 per cent, the par valae being $10Q
per share?
Sdutioiu
$1100 BB par ralae of 11 shares.
66 ^ 6 per cent discount
$1034 tss real yaluc, or required cost
8. How mach will 53 shares Suffolk Bank stock cost at an
advance of 23 per cent^the per Talue being $100 per share ?
4. How much will 43 shares Vermont Central Railroad
stock cost, at a discount of 78 per cent, the par value being
$50 per share?
5. How many shares of stock at an advance of 5 per cent
on the par value of $100 per share, can be bought for $1995 ?
6. How many shares of stock at a discount of 5 per cent
from the par value of $100 per share, can be bought for
$1805 ?
7. A broker paid $1776.50 for stock at an advance of 4^
per cent. What was the par value of the stock bought ?
8. A broker paid $4850 for bank stock, at a discount of 3
per cent. What was the nominal value of the stock bought ?
9. A broker sold on consignment 376 shares bank stock,
par value $100 per share, at an advance of 7 per cent. He
charged 25 cents per share for his services. How much
ought he to remit to the person consigning the stock ?
10. I bought 40 shares of railroad stock, at 33 per cent be-
low par, and after keeping them 10 months, sold them at iiO
per cent below. How much did I gain on them, allowing
that I hired money for the investment at 6 per cent interest,
and that the par value of the shares was $100 each ?
11. Alfred £. Potter bought 10 shares of bank stock at
par, which was $50 per share. At the end of 3 months he
received a dividend of 4 per cent, and at the end of 9 months
he received another of 3 per cent. At the end of one year
he sold the stock at an advance of 3 per cent Money being
worth 6 per cent per year, how much did he gain by tYiQ
transactions 1
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INSURANCB. 909
NoT» — Interest should be reckoned on the dividends from the tiiiM
Jicy were made.
12. Crockc: & Guild sent $972.63 to a stock broker,
directing him t:o invest it in Fall River Railroad stock. He
bought the stock at a premium of 7 per cent, the par vaiae
being $100 per sliare, and he charged a commission of 1 per
cent on the money invested. How many shares did he buy ?
13. I paid $7398 for stock at 10 per cent below par, and
some time after sold the stock at 10 per cent above par. How
much did I gain on it ?
14. Mr. Hamblin bought stock at 10 per cent above par,
bat he was obliged to sell it at 10 per cent below par. Allow
iDg that he lost $138.40 on it, what did he pay for it ?
30 4 • Insurance.
(a.) Insttbance is an obligation assumed by one individv <t
or company to pay to another a certain sum of money on 0,e
occurrence of some contingent event.
(6.) A honse is insured against loss by fire, when some indiyidnal or
company agrees to pay the owner a specified sum if it is burned within •
given time.
(c.) A ship in like manner may be insured against loss by fire or by
any of the perils of the sea.
(d.) When the obligation is to pay the person insured a certain sum
if ho is sick, it is called health insurance ; when it is to pay to his
heirs, or to some particular person, a specified sum if the person insured
dies, it is called life insurance.
(c.) The written certificate of insurance is called a policy.
{/.) The sum paid for insurance is called the premium.
(g.) If property is insured, the premium is a certain per cent of the
tnm covered, i. e., of the sum for which it is insured, and is usually paid
at the time of effecting the insurance.
(A.) When health or life is insured, the premium Is usually a given
lum paid annually during the time for which the insurance continues;
and its amount is calculated from tables prepared by the different insur-
ance companies. It will, therefore, be unnecessary to give examples in
life or health insurance.
1. What would be the expense of getting $1000 insured
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510 ASSESSMKNT OF TAXES.
on a house at a premium of 1^ per cent, the charge for tha
policy being $1 ?
2. I had $1000 insured on my house for 7 years, for which
I paid a premium of 2^ per cent, and $1 for the policj. At
the end of 5 jrs. 7 mo. 15 da. the house was burned, and I
received the amount for which it was insured. Money being
worth 6 per cent per year compound interest, how much
did I really save by ha ring the house insured ?
8. Oct 1, 1854, 1 bought a lot of flour for $6000, giving
my note payable in 6 months, and immediately shipped it for
England. Oct 8, 1854, 1 got it insured for $6000, paying a
cash premium of Itf^ per cent, and $1 for the policy. Oct 10,
1854, 1 paid a bill of $50 for cartage and other expenses.
Nov. 1, 1854, I received information that the vessel was lost
at sea; and Dec 1, the insurance company paid me the in-
surance. I immediately put it on interest at 6 per cert, till
my note for the flour became due, when I made a complete
settlement Had I gained or lost by the transactions, and
how much ?
30tS« Assessment of Taxes,
(a.) Taxes are duties or assessments laid on persons or
property, usually for some public purpose.
(b,) A tax on persons is called a poll tax, and a tax on
property is called a property tax.
The poll tax is only assessed on males between certain ages, as he>
tween twenty-one and seventy ; and in some states is not assessed
(c.) In assessing taxes, it is necessary, —
First. To estimate the value of all the property to bo
taxed, and make a complete inventory of it.
Second. To find the number of polls, i. e., the number of
persons liable to pay a poll tax.
Third. To determine what portion of the tax is to be
raised upon the polls, and to divide it equally among them-
Fourth. To find how much must be paid on each dollar of
ilie taxable property, to raise the remainder of the tax
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ORDERS. 311
Thif last may be done by dividing the amount to be raised by the
estimated value of the property on which it is to be raised. It will thca
be easy to find the tax of any individual.
1. A tax of $4800 is to be raised hy a certain town* Tho
taxable property is valued at $960,000, and there are 320
polls, each taxed $1.50. What will be the tax on each dol-
lar, and what will be the tax of each of the foUowicg persons ?
A, who pays a tax on $5700, and 2 polls.
B, who pays a tax on $728, and 1 poll.
C, who pays a tax on $8976, and 3 polls.
D, who pays a tax on $1147, and 1 poU.
Solution, — The tax on 320 polls at $1.50 each is $480, which, sub-
tracted from $4800, leaves $4320 to be assessed on the property.
Since 960,000 is to be taxed $4320, one dollar will be taxed flflol i ao y
of $4320, which is 4 j- mills.
A's tax on 2 polls would be twice $1 .50, or $3, and on $5700 prop-
erty would be 5700 times 4^ mills, which is $25.65, and added to $9
gives $28.65 as the amount of A*s tax.
The tax of the others may be found in the same way.
2. A tax of $4800 is to be raised in a certain town. T*ie
taxable property is valued at $1,228,000, and there are 575
polls to be taxed $1.40 each. How much is the tax on $1 ?
How much is the tax on each of the following persons ?
J. S., who pays a tax on $1728, and 1 poll.
S. B., who pays a tax on $4235, and 2 polls.
L. M., who pays a tax on $7945, and no polL
F. G., who pays a tax on $2794, and 1 polL
»06. Orders.
(a.) If a person should wish to have money which is dud
him paid to some one else, he might write an order for it,
i. e., a paper requesting the one who owes it to pay it to the
third person.
(&.) Suppose, for instance, that Edward Mellen owes Bichard Jack-
son fiye hundred dollars, and that Mr. Jackson owes Stephtn Baker
one hundred dollars.
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Sit OBDBB8.
(e.) Soppofe thmt ICr. Baker presents his bill ibr payment, and tha
Mr. Jackson, not baring the money by him, ^res him the following
•ider on Mr. Mellen :— -
vCea4« Itaa to lite oidel oC u te|Xeiv lSSa£«%
•MA luifi^Ud doEial^i oftd cnaVae ln«. Kune t» m«.
(dl) In considering this order, we may notice, —
First The ** $100 " and the date at iu head. See X%% (d)
• Second. ** Edward Mellen, Esq.," which is written to show to whom
the order is addressed.
Third. The request, " Please pay to the order of Stephen Baker one
bnndred dollars.** The explanations of ISSy (d.) will show the nse of
tfie rarions parts of this.
Foorth. ** And charge the same to me,** which authorizes Mr. Mellen
to chaige the money to Mr. Jackson, as though it had been paid directly
to him. The phrase " on my account ** would mean the same thing.
(e.) This order, when written, would be taken to Mr. Mellen by Mi
Baker. If Mr. Mellen pays it, Mr. Baker indorses it, or writes the
words " Beceired payment,** with his name, upon it, and hands it to Mr.
Mellen, who keeps it as evidence that he has actually paid out so much
money on account of Mr. Jackson.
(/) By this arrangement it will be seen that Mr. Jackson pays the
one hundred dollars which he owes to Mr. Baker, and that Mr. Mellen
pays one hundred dollars of his indebtedness to Mr. Jackson.
{g.) If Mr. Mellen should refuse to pay the above order, Mr. Baker
wonld have no right to commence legal proceedings against Aim, but he
would have the same claim against Mr. Jackson that he had before the
order was given.
(h,) If Mr. Mellen is willing to pay it, but is unable to do so the day
that it is presented, he should accept it, by writing his name, or the word
** accepted ** and his name, upon it This is called an acceptance^ and
wonld give Mr. Baker the same legal claims against Mr. Mellen that he
would have had upon a note for the same amount
(i.) An order may be given to pay in a given time, as '* thir^ dayi
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BILLS OF EXCHANGB^ 819
after date." Snch an order should at once be presented for acceptance,
80 that the person- on whom it is drawn may be held responsible for itf
payment.
(j.) An order may be payable in some given time after tight^ i. o., in
some given time after it is exhibited to and accepted by the person (o
whom it is directed. Usage varies nrith regard to grace on soch orden^
thongh it is commonly allowed.
207. Bilh of Exchange.
(a) When an order is drawn on a person living in a distant plaot. it
Is caUed a bill of exchange.
(h.) Bills of exchange are of two kinds — foreign and inland; the
former being drawn on persons living in foreign countries, and the latter
on those living within our own. There are, however, no other essential
differences between foreign and inland bills, than such as are connected
with the different business customs existing in different countries.
(c.) It is customary to write two or three copies of a foreign bill of
exchange, and to send them by different vessels, so as to render it
reasonably certain that one of them shall reach the person to whom it is
sent These bills constitute a set of exchakoe, and are called the
first, second, and third of exchange. They are so worded as to express
that the payment of one renders the others void.
(d.) The following illustrates theur form : —
SgocR^cutoe poV mIOOO*
cTu^enAuf daa6 a&e^ Mohi oE tfiX^- UAAl of edccfuutoey
^Secofvd futa wuM un|bala^ [utu^ Y» toe oVoeV oE cSiouHt cwta
cSiUEei oit» laou^^arto }mwm%'^ aito cfuxAAe \m, 4<uite ko o»i
a^cooiuit. xPotUv ^ affanuifreiio.
{%,) The other bills of the set would be dated and directed like tlii%
bat the second wonld read, —
27
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114 BILLS OF EXCHANOK.
** Twentf dayi after tight of this second of exdiange, (firs, sod Ak4
vspaid,) pajr," &c^ as before.
A corresponding change wonid be made in the third.
(/) To iUnstrate the nse of such bills, let ns suppose that Brown
and Butler, of Boston, owe Miller and Jones, of Liverpool, one thon-
•atid pounds, and that French, Harris, & Co., of Liverpool, owe Poitet
and Hammond, of Boston, one thousand pounds.
{g.) It is obrious that if the parties should pajr their debts in speciej
they must incur the expense and risk of transporting two thousand
pounds across the Atlantic, and that then as much money would h&ve
been brought back to each country as has been sent out from it.
(k) But if Brown and Butler should buy of Potter and Hammond,
a set of exchange for one thousand pounds on French, Harris, & Co^
and should indorse it to Miller and Jones, and send it to them, and they
should collect it, all these debts would be cancelled without any trans-
fer of specie from one country to the other.
(i.) For Potter and Hammond would have received from Brown and
Butler an equivalent for their claim against French, Harris, & Co^ and
have thus got the means of paying Miller and Jones. Miller and Jones
would have received from French, Harris, & Co. an equivalent for their
claim against Brown and Butler ; and hence each would have paid out
the value of what he owed, and have received the value of what was
due to him.
(;.) Such transactions as the above are so manifestly to the advan-
tage of the parties concerned in them, that a demand for bills of ex-
change would naturally spring up in countries having much business
intercourse with each other, as between the United States and England,
or the United States and France.
(k.) If our merchants should owe the merchants of England more
than they owe us, there would be more persons here wishing to buy than
tell bills on England. The demand would thus be greater than the
supply, and bills would bring more than the sum for which they were
drawn, or would be at a premium,
(/.) If, however, English merchants should owe ns more than we owe
them, more of our merchants would wish to sell than to buy, and bills
on England would be at a discount >
(m.) There would be a limit to the premium on one side, and to the
discount on the other ; for merchants would not pay more premium for
bills of exchange than it would cost tliem for freight and insurance to
transport specie across the ocean.
(n.) When exchange on a foreign country is at a premium, it is said
•o be against us, for it not only makes our merchants pay more than the
amount of their debts, but it indicates that a balance of specie is due to
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BILLS OP EXCHANGE. 815
that cooutry. When it is at a discount, it is said to be in onr favor, fcf
it not only enables our merchants to pay their debts for less than theii
amonnt, but indicates that a balance of specie is due to us.
{o.) These variations in value are called the course of EXcnANoa.
(;/.) It is an important thing with the merchant to watch the course
of exchange between different countries, as he can often save consider-
able money by making his remittances through an indirect route. Thus,
it may happen that bills on England command a much higher premium
than bills on France ; at the same time it may happen that exchange
fiom Franco on England can be bought at a lowj^ate. In such circum-
stances it may happen th<it by first buying a bill on France, and then
Knding it to a banking house there, with directions to buy a bill on
England, a merchant may pay his debt for less than he could by direct
remittance.
(q,) The pound sterling varies, as has been said, from $4.83 to $4.86,
but in exchange its par value is assumed to be $4.441- ; so that when a
pound of exchange brings its true value, it should be very nearly 9 per
cent above par.
1. Paine & Colman, of Providence, bought of Bobert
Greene & Co. a set of exchange for £1500 on R. & S. Hub-
bard, of Liverpool, paying 10 per cent premium. How many
dollars did it cost them ?
2. What will a set of exchange on London for £2000 cost
at 9 J per cent premium ?
3. What will a set of exchange on France for 5000 francs
cost at 1 per cent discount, the par value of the franc being
$.186 ?
4. Which will be the most profitable, and how much the
mobt, to buy a set of exchange on London for £1500 at 10^
per cent premium, or to buy sovereigns enough at $4.86 to
pay the debt, allowing that it will cost 2^ per cent of their
value for freight and insurance ?
5. How much will it cost to remit £3000 to Liverjiool
tijrough France, when exchange on France can be bought at
the nite of 5.4 francs for $1, and exchange on England can
be bought in France at the rate of £1 for 24 francs, allowing
that a banking house in Paris will charge ^ per cent for pur-
chaoing the exchange in that c tjr ?
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Ji6 PBOFrr AND LOM.
308. Profit and Loss.
Mo6t problems in profit and loss come under one of thre*
cWscs, viz. : —
First Those in which it is required to find for what price
articles must be sold that th3ir owner maj gain or lo3e a cei^
tain per cent of their cost
Second. Those in which it is required to find what per
cent of the given cost will b3 gained or lost by selling articles
at a given price.
Third. Those in which it is required to find the cost, cnt
some per cent of the cost, of articles on which a certain per
cent will be gained or lost by selling them at a given price.
They involve similar principles to those involved in other
examples in percentage and fi*actions.
1. A speculator bought a lot of land for $2473. For how
much must he sell it to gain 25 per cent of its cost ?
Suggestion, — He must sell it for the cost, plus 25 per cent of the
cost ; or, sinco 25 per cent =» ^, he mast sell it for the cost, plus ^ of
the cost*
2. Mr. Huntington bought a lot of grain at 54 cents per
bushel ; but it being damaged, he was obliged to sell it at a
loss of 16| per cent For how much per bushel did he
sell it?
Suggesliott, — Since isf per cent = i, he must have sold it for the
fX)st, minus \ of the cost*
3. Mr. Shelley bought a lot of sugar at 8 cents per pound.
For how much per pound must he sell it to gain 12|. per
cent?
4. For how much per yard must cloth, costing $2.50 per
yard, be sold, to gain 10 per cent on its cost?
* When the given per cent can be reduced to a convenient vulgar frao-
tton^ as in this example, it is ordiu'jrily best to use the vulgar |rai:tioa
Inttoad of the given per cent
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PROFIT AXD LOSS. 317
5. 1 bought cloth at $2 per yard, and sold it at $2.33 per
jrard. What per cent of its cost did I gain ?
Solutim. — Since I gave $2 per yard for the cloth, and received
$2.33, 1 received $.33 per yard more than I gave ; and as $.33 ss= j^^
of $2, my gain must be ^^ = 16^ per cent of the cost
6. What per cent shall I lose by selling molasses, which
tost me $.30 per gallon, for $.25 per gallon ?
Solution. — Since I bought the molasses for 30 cents, and sold it for
S& cents, per gallon, I lost 5 cents per gallon on it ; and, as 5 cents sa
^ = i of 30 cents, I lost i, or icf pe: cent of the cost.
7. How much per cent shall I gain by selling floor, which
cost me $6.50 per barrel, for $7,215 per barrel ?
8. Mr. Winsor bought flour at $7.50, and sold it at $6.75
per barrel. How much per cent did he lose ?
9. What per cent shall I gain by selling linen cloth, which
cost 45 cents per yard, at 50 cents per yard ?
10. Whit per cent shall I lose by selling linen cloth, cost-
ing 50 cents per yard, at 45 cents per yard ?
11. A merchant bought some molasses at $.20 per gallon,
and some oil at $1.16 per gallon. He sold the molasses at
$.25 per gallon, and the oil at such rate that he gained the
same per cent on it that he gained on the molasses. For
how much per gallon did he sell the oil ?
12. I gained $17.28 by selling a lot of sugar for 16 per
cent more than it cost. How many dollars did it jsost, and for
how many did I sell it ?
Suggestion. — Since I gained 16 per cent, or 2*5- of the cost, I must
have sold it for §|- of the cost. Hence, the cost = ^^, and the price
for which I sold it == -^ of the gain.
13. Mhp Kent bought a lot of apples, and sold them for 20
per cent more than they cost, by wfcich he gained $24.80.
How much did they cost him, and for how much did he sell
them?
14. Mr. Kilbum sold 43 barrels of apples for $6.45 less
fhan they cost him, thereby losing 10 per :ent of their cost
ait
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dl8 PROFIT AND L088.
What did they cost him, and what did he get per barrel fbi
them?
15. Mr. Thurbur sold 14G yards of cloth for $71.54 more
than it cost him, thereby gaining 14 per cent. How much
did he receive per yard for it ?
* 1 6. Logee dc Drown sold a large lot of goods for $6700.43 J,
thereby gaining 18 per cent on their cost How much did
the goods cost them ?
Suggestion. — Since they gained 18 per cent of the cost, $6700.43^
must be 1 18 per cent, or | j of the cost
17. What is the cost of a lot of goods on which 15 per
cent will be gained by selling them for $288.65 ?
18. What is the cost of a lot of goods on which 8 per cent
will be piined by selling them for $622,215 ?
19. What is the cost of a lot of goods on which 7 per cent
will be lost by selling them for $442.68 ?
Suggestion. — Since 7 per cent will be lost by the sale, $442.68 mast
•qnal 93 per cent of the cost
20. What is the cost of a lot of goods on which 30 per
i 3nt wiU be lost by selling them for $874,846 ?
21. What is the cost of a lot of goods on which 9 per cent
'.rill be lost by selling them for $9009 ?
22. A man bought com at 50 cents per bushel, for which
he asked 25 per cent more than it cost him ; but it falling in
price, he was obliged to sell it for 25 per cent less than his
asking pric5. Did he gain or lose, and how much per cent ?
How much on each bushel ?
Solution. — Since his asking price was 125 per cent, or J of the cost,
and his selling price was 75 per cent, or f of his asking price, his selling
price must have been | of ^, or {f of the cost Therefore he lost iV?
or 6^ per cent of the cost ; and since each bushel cost $.5(J^his loss per
bushel must have been t^q of $.50, which is $.03^.
23. A merchant asked for a lot of goods 12J per cent more
tlian they cost him, but was obliged to deduct 12^ per cent
from his asking price. What part of the cost did he lose ?
24. I asked for a lot of 4otb JO per penf more thfm it opal
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PBOFIT AND LOSS. 819
ID 3, but was obliged to deduct 10 per cent fix)in my asking
pnce. What per cent of its cost did I lose ?
25. A watch dealer bought a watch for $75, and asked for
it 33 J per cent more than it cost. lie was obliged to sell it
for 10 per cent less than his asking price. What per cent did
he gain on the investment ? How many dollars ?
2G. A merchant asked for a quantity of goods 25 per cent
more than they cost him, but was obliged to sell them for 1 2^
per cent less than his asking price. lie gained $98.70 by the
transaction. How much did the goods cost ? For how much
did he sell them ? What was his asking price ?
Suggestion. — His asking price was { of the cost, and his felling pHc«
was j- of his asking price, or j- of f <» -J-i^ of the cost. Ueoco, the
gain, $98.70, = i^ of the cost.
27. A merchant asked for a lot of cloth 20 per cent more
than it cost him ; but being in nant of money, ho told the
cloth for 25 per cent less than his asking price. Allowing
that he lost $.275 per yard, for how much per yard did he
sell it?
28. I sold a lot of goods for $42G8, which was f^^ per cenfc
less than my asking price. My asking price wa8 44 per cent
more than the cost of the goods. I low much did they cost ?
29. By selling flour at $G.G5 per barrel, I ijhall lose 5 per
eent of its cost For bow much per barrel must 1 sell it to
gain 5 per cent ?
Suggestion. — $6.65 = 95 per cent of the cost Hence, 105 per cent
of the cost mast equal ^^ of $6 65.
30. By selling land at $88.14 per acre I shall gain 13 per
cent For how much must I sell it to gain 20 per cent ?
31. By selling a lot of goods for 81 13.75 I lost 9 per cent
For how much ougfit I to have sold it to gain 9 per cent?
32. I bought 400 buslieli of grain, and sold half of it at
75 cents per bushel, which was 20 per cent more tlian it coit
I sold the remainder at an ailvance of 25 [j^r cent on the cost
For how much per bushel did I sell the last lot ? Uow many
dollars did I gain ?
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ISO PBOFIT AND LOSS.
83. What must be the asking price of cloth costing $2.5S
per yard, that I maj fall 10 per cent on it, and still gain 14
per cent on the cost ?
Su'jfftgtion, — As the selling price is to be 90 per cent, or Tfiy% of th«
Mking price, and 114 per cent of the cost, the asking price most be-V^
of the lelUng price, and "VV^ of {ii = -^ of the cost
84. What must be the asking price of cloth costing $3.29
per yard, that I may deduct 12j^ per cent from it, and still
gain 12^ per cent on the cost ?
85. What must be the asking ) rice of boots costing $2.75
per pair, that I may fall 16 J per cent on it, and still gain 20
per cent on their cost ?
86. If, by selling cloth at $2.34 per yard, I gain 4 per cent
of its cost, what per cent shall I gain by selling it at $2.79
per yard ?
Suggestion and Answe*, — Since $2^4 = 4 per cent more than the
:o8t, it must equal 104 per cent of the cost. Hence, $2.79 = § J| of
104 per cent, = 124 per cent of tho cost, and the gain equals 24 per
cent
87. If, by selling cloth at $4.37 per yard, I gain 15 per
cent of its cost, what per cent shall I gain by selling it at
$4.94 per yard ?
38. If, by selling flour at $6.75 per barrel, I gain 8 per
cent of its cost, what per cent shall I lose by selling it at $6
per barrel ?
39. If, by selling oil at $1,254 per gallon, I lose 5 per cent
of its cost, what per cent shall I gain by selling it at $1,584
per gallon ?
40. I bought 96 acres of land at $84 per acre, and sold }
of it for the cost of the whole. What per cent did I gain on
the part sold ?
Suggestion. — Tlic part sold cost f , and was sold for } of the cost of
the entire lot. Hence, it was sold for ^ of its cost
41 . I bought 2^tons of iron at $48 per ton, and sold f of
it for the cost of the whole. Wl at per cent did I gain on tha
Dart sold ?
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PBOFIT AND LOSS. 521
42. I bought 83 barrels of beef at $12.50 per barrel, and
was obliged to sell it for what f of it cost. What per cent
did I lose ?
43. A. & M. J. Miles bought of Cragin & Cleveland, for
cash, goods to the amount of $423.75, and the same daj sold
them at an advance of 16 per cent, receiving in payment a
note on 3 months. This note they got discounted at a bank at
the rate of G per cent per year. How much did they gain on
the goods ?
44. Armington, Horswell, & Kilburn bought of Godding,
Briggs, & Co. goods to the amount of $1000, payable in 6
months, without grace. One month afterwards they sold the
goods for caoh, at an advance of 10 per cent, and immediately
put the money at interest at 6 per cent. When the 6 months
had expired, they collected the amount of the money they had
lent, and paid the bill due Godding, Briggs, & Co. Did they
gain or lose, and how many dollars ?
45. I bought a lot of coffee at 12 cents per pound. Allow-
uig that 5 per cent of the coffee will waste in weighing it out,
and that 10 per cent of the sales will be bad debts, for how
much per pound must I sell it to make a clear gain of 14 per
cent on the cost ?
Answer, 16 cents per ponnd.
46. What must be the asking price of raisins costing $7.29
per cask, that I may fall 10 per cent from it and still gain 10
per cent on the cost, allowing that 10 per cent of the sales
will be bad debts ?
47. I bought 8 casks of oil, each containing 133 gallons,
at $1.20 per gallon, and paid $5.32 for having it brought to
my store. Allowing that there will be a waste of 5 per cent
in measuring, that 3 per cent of my sales will be bad debts,
and that it will cost 1 per cent of the remainder to collect it,
for how much per gallon must I sell it to make a net gain of
83 per cent on its cost al my store, nothing being allowed fo»
interest ?
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322 PAUTNF.RSHIP.
309. Partnership.
Twc or more persons, uniting for the purpose of ca..TyiDg
ou business together, form wliat is called a pabtnershiP;
riHM, or COMPANY. The capital invested by them is called
their stock in tbadb.
It is erideot that iho profit or loss made by the company should be
shared among iu members in proportion to what the use, or interest, of
each man's stock for the time it was invested is worth.
When the stocks of the several partners are invested for the same
length of time, their use, or interest, will be proportioned to the stock?
themselves, and hence each partner's gain or loss will be the same part
cf his stock that the entire gain or loss is of the entii*e stock ; or it will
be the same part of the entire gain or loss that his sfock is of the entire
stock.
1. A, B, and C trado in company. A puts in $250, B
puts in S750, and C puts in $500. At the end of 6 months
they find that they have gained $472.50. What is each man's
sliare of the gain ?
First Solution. — Since A's stock = $250, B's = $750, and C's =a
$500, the entire stock = $250 -+- $750 -|- $500 = $1500; and as the
gain = $472.50, it must equal -^/jn^n, or jpj^ of the stock. There-
fore, each man's gain will be ^:^ of his stock, which gives for A'a
gain $78.75, for B's gain $236.25, and for C's gain $157.50.
Second Solution. — Since A's stock = $250, B's = $750, and C's =
$500, the entire stock must equal $1500, of. which A's stock = T^^^y
= i» B's = ^:Pjj = i, C's = i^^'V == i. Therefore, A should have
i, B should have i, and C should have ^ of the gain, i of $472.50
= $78.75 = A's share; ^ of $472.50 = $236.25 = B's share; i of
$472.50 = $157.50 = C's share.
2. X, Y, and Z traded in company for 1 year. X put in
$1000, Y put in $1500, and Z put in $2000. At the end of
t,he year they found that they had gained $1800. What was
each man's share of the gain ?
3. A man, failing in business, finds that he owes A $424,
B $638, C $197, D $338, and E $574, and that his whole
available property amounts only to $1173. How much ought
he to pay to each creditor ?
\
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PARTNERSHIP. 323
Suggestion. — Sinc^ he owes $2171, and has but $1173, he can pa^
but iilf of his debts. Therefore, he ought to pay A ^}fj of $424,
B if? f of $638, &c
4. The stock of a bankrupt is valued at $1200, and he
owes $4200. How many dollars ought he to pay the person
to whom he owes $546 ? to whom he owes $338.73 ?
5. A, B, C, and D agree to cut 500 cords of wood for
$300. When the job is finished, they find that A has ciit 125
cords, B 100 cords, C 150 cords, and D the rest. How many
dollars ought each to receive ?
6. A and B traded in company. A put in $200, and B
put in $300. A's share of the gain was $84.56. What was
B's share ?
7. A and B traded in company, and gained $348, of which
B's share was $261. If A's stock was $175, what was B's
Rtock, and A's share of the gain ?
8. Samuel Greene and Joseph Irons traded in company.
Greene paid in 3 times as much of the stock as Irons, and
they gained $1176. What was each one's share of the gain ?
Suggestion, — Since Greene paid in 3 times as much as Irons, both,
together must have paid in 4 times as much as Irons. Therefore, Irons
paid in j-, and Greene £ of the stock.
9. William Balch and Joseph Adams bought a ship to-
gether, Balch paying in twice as much money as Adams. At
the end of one year they sold her, and found that they had
realized a profit of $15,000 from her. What was each part-
ner's share ?
10. Anderson and Parker, after trading in company for 2
years, found that their profits had been $2400. Allowing
that Anderson's stock was J of Parker's, how many dollars
of the profit ought each to have ?
11. A, B, and traded in company. A put in -J of the
stock, B put in ^ of it, and C put in the rest. On dividing
tlie gain, they found that C's share of it was $321. What
was the gain of each of the other partners ?
12. William Hall, Edward Johnson, and Henry Whiting
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824 PABTNKR8HIP ON TIMB.
tnded in company, and gained $6534, of which JohnscHr**
share was $1089. If Johnson's and Whiting's stock was
' together equal to twice Hall's, what was Hall's share of the
gain ? What was Johnson's share ?
13. A small estate belonged to a large number of heirs :
2 members of the family of A each owned y^ of the estate ;
4 of the famOy of B each owned ^ of it ; 4 of the family
of C each owned -j^ of it ; 2 of the family of D each owned
^\f of it ; 4 of the family of E each owned ^^ of it ; 3 of the
&mily of F each owned -^ of it ; 4 of the family of G each
own^ ^f^ of it ; 6 of the family of H each owned y§t ^^ ^^*
8 of the family of I each owned ^ of it. Mr. Byram, as
agent for the above-named individuals, sold their interest in
the estate for $350. How many dollars ought he to give to
each?
Ansivar.
$ 3.123 to each member of A*8 family.
$ 1.562 to each member of B's family.
$31,234 to each member of C's family.
$15,617 to each member of D*8 fkmilj.
$12,493 to each member of E^s family.
$ 1.785 to each member of F*s family.
$ 2.499 to each member of G*8 family.
$ 8.924 to each member of H's family.
$20,823 to each member of Vb fomily.
KoTE. — The above example is a statement of transactions which
actnally occnrred. It was broaght to the anthor for solution, by th«
agent of the parties.
210. Partnership on Time,
In dividing the gain or loss among the partners, when their
shares of the stock are invested for unegaal times, it becomes
necessary to coq^ider both the stock and the time, or to con«
sider the interest of each man's stock for the time it was in
trade. The following examples and solutions will illustrate
this: —
1 A, By and C traded in company. A put in $750 for 10
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PARTNERSHIP ON TIMR. 825
joioiUhs, B put in $375 for 12 months, and C put in $1125
Ibr 16 months. Thej gained $860. What was each man's
rkare of the gain ?
First Solution,
$37.50 = interest of A's stock for 10 mo.
22.50 = interest of B's stock for 12 mo.
90.00 = interest of C's stock for 16 mo.
$150.00 = interest of whole.
Therefore, A should have AWV. or J, of the gain, = $215.
B should have -^to^* or ^^, of the gain, = $129.
C should have AWi7, or f, of the gain, = $516.
Second Solution,
The use of $ 750 for 10 mo. is worth the use of $ 7500 for 1 mo.
The use of $ 375 for 12 mo. is worth the use of $ 4500 for 1 mo.
The use of $1125 for 16 mo. is worth the use of $18000 for 1 mo.
Use of whole stock is worth the use of $30000 for 1 mo.
Therefore, A should have f^T^js^y, or i, of the gain, = $215.
B should have ^VirV, or ^, of the gain, = $129.
C should have i§8^^» or f , of the gain, = $516.
When the stocks of the several partners are convenient multiplet
or fractional parts of each other, a very neat solution can be given.
Thus, in. the above example, bj noticing that B's stock equals i of A's,
and that CTs stock equals ^ of A's, we maj have the following : —
Third Solution. — The use of A's stock 10 mo. = use of 10 times
A*s stock for 1 mo.
The use of B's, or j- of A's stock, 12 mo. = use of -^, or 6 times
A's stock for 1 mo.
The use of C's, or f of A's stot k, 16 mo. == use of ^, or 24 times
A's stock for 1 mo.
Use of whole = use of 10 + 6 + 24, or 40 times A's stock for
1 mo.
Therefore A should have iJ, or J ; B yV or ^ ; and C ft, or f ,
of the gain, which will give the same answer as before.
2. Charles French, Francis Baker, and Otis Atherlon
traded in company, under the name of Charles French & Ci
French put in $1000 for 20 mo., Baker put in $800 for 16
mo., and Atherton put in $500 for 20 mo. They gained
$1500. How many dollars of the gain ought each to receive ?
28
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82C PABTNEBSatP ox TIMK.
8. George Jackson, William Leach, and Albert Buflin^i^toQ
traded in company. Jack>on put in $144 for 6 mo., L><.*ach
put in $72 for 7 mo., and Buflington put in $21 C for G mo.
20 da. Tbej gained $114. What was each man*s share of
the gain ?
4. A, B, C, and D hired a pasture together, in which A
pastured 4 cows 13 weeks, B pastured 5 cows 16 weeks, C
pastured 8 cows 10 j- weeks, and D pastured 4 cows 16 weeks.
The rent of the pasture was $102. How many dollars ought
each man to pay ?
5. Samuel Austin, Jacob Brown, and Moses Sumner
formed a partnership for 2 years, under the name of Samuel
Austin 6c Co. Austin at first paid in to the stock $1000, but
after 8 mo. had elapsed he paid in $500 more. Brown at
first paid in $1250, and 16 mo. afterwards he paid in $250
more. Sumner at first paid in $1500, but at the end of 16
mo. he took out $500. They gained $3600. What was each
man's share of the gain ?
Solution,
Interest of SIOOO for 8 mo. = « <» I = $i6o = 5„t Anstm's stoct
Interest of $1500 for 16 mo. = $120 )
Interest of $1250 for 16 mo. = $100 j _ _ .
Interest of $1 500 for 8 mo. = $ 60 ) *
Interest of $1500 foH6 mo. = $120 ) ^,^^ .,« _, ^ ,.
T * * /><»i/vrv/^r o • .« }" =$160s=mt. Sumner's stock.
Interest of $1000 for 8 mo. ss $ 40 j
By this, it appears that the interests of their respective stocks, for the
time they were in trade, were alike. Hence, the gain should be divided
equally, and each partner should have i of $3600, which is $1200.
KoTE. — Other solutions similar in character to those given to the
first example might have been added j but as the pupil can readily dis«
cover them, they have been omitted.
6. Joseph Southwick, Francis Lowe, and Henry Taft
formed a partnership for 3 years, under the name of Soufli-
wick, Lowe, & Taft. When they commenced business, each
partner put in $3000 ; but at the end of the first year South-
wick put in $3000 more, and Lowe withdrew $1500. At the
end of the second year, Southwick withdrew $2000, and
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POWERS AND ROOTS. 827
Lowe put in $4000, and Taft put in S2000. Wien the part-
nership expired, they found that they had gained $9000.
What was each partner's share of the gain ?
7. S. Gamwell, C. Grover, R. Wheelock, and W. Godding
formed a partnership, under the title of Gamwell, Grover, &
Co. Gamwell at first put in $8000, but at the end of 6 mo.
he withdrew $2000, and at the end of 12 mo. he withdrew
$1000 more. Grover at first put in $G000, but at the end
of 10 mo. he put m $3000 more. Wheelock put in $7000.
Godding at first put in $10,000 ; at the end of 6 mo. he with-
drew $2000, and at the end of 14 mo. he put in $4000. At
the end of 2 years they found that they had gained $12,000.
What was each man's share of the gain ?
SECTION XVI.
POWERS AND ROOl'S
2 1 1 • Definitions.
{a,) The product of a number taken any number of times
as a factor is called a power of the number. — See \0S^
(d.) (c.) (/.) and Note.
(^.) A ROOT of any number is such a number as, taken
some number of times as a factor, will produce the given
number.
{c.) If the root must be taken twice as a factor tp produce
the number, it is the square root, or the second root ;
if three times, it is the cube root, or the third root ; if
four times, it is the fourth root ; &c
Thus, 2 is the square root of 4, the third root cf 8, the fourth root of
16, &c^ because 22 = 4, 2^ == 8, 2* = 16 ; Lq,
(d.) The character ^, called the radical sign, is used
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128 BBLATION or ▲ 8QUABE TO ITS ROOT.
to indicate that the root of the number over which it is placed
is to be extracted.
(e.) The degree of the root is indicated bj a small fig-
are, called an index, which is placed a little above and at the
left of the sign. When no index is written, the square root
is required.
Thus, ^ 4, or ^ 4, means tlie square root of 4.
i/'243 means the fifth root of 243.
i/ 7^ means the fourth root of the 3d power of 7.
(/.) We may also indicate that a root is to be extracted,
by using a fractional exponent
Thus,9i = y9; (125)* =»^"l25j 27ts=^27«; &c
(ff,) The process of finding the powers of numbers is
called Involution, and the process of finding their roots is
called Evolution, or the Extracting of Roots.
913* jRelation which the Denominations of a Square hear
to those of its Root.
(a.) TABLE OF SQUARES.
1«==1 4«=16 72 = 49
22 = 4 52 = 25 82=64
32 = 9 62 = 36 92 = 81
109 -= 100 10,0002 = 1.000,000,000
1002 — 10,000 100,0002 = 10,000,000,000
1,0002 — 1,000,000 1,000,0002 = 1,000,000,000,000
(6.) The above table shows that, — First There are below 100 bnt 9
entire numbers which are perfect squares.
Second. The entire part of the square root of any number below 100
will be less than 10, and therefore contain but 1 figure; of any num-
ber between 100 and 10,000 will lie between 10 and 100, and therefore
contain 2 figures; between 10,000 and 1,000,000 will lie between 100
and 1000, and therefore will contain 3 figures ; &c.
1. How many figures are there in the entire part of the
square root of 865698 ?
Antwer. — Since 865698 lies between 10,000 and 1,000,000, its root
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DIVISION INTO PERIODS. 329
must lie between the roots of those numbers, i. e., between ICX) and 1000,
and must therefore contain 3 figures in its entire part.
How many figures are there in the entire part of the
root of —
2. 69748769? I 4. 12496743297?
3. 486497950068? 5. 5847695329?
313* Division into Periods,
(a.) As the square of 10 is 100, of 100 is 10,000, &c., it follows that
t]i8 square of any number of tons will be some number of hundreds ;
of any number of hundreds will be some number of ten thousands, &c ;
or, in other words, that the square of tens will give units of no denomi-
nation below hundreds ; the square of hundreds will give units of no
denomination below ten-thousands ; &c.
(6.) Hence, the two right hand figures of any number will contain
no part of the square of the denominations of the root above units ; the
four right hand figures will contain no part of the square of those above
tens, &c.
(c.) Therefore, if we should begin at the right of any number, and
separate it into periods of two figures each, the number of periods would
be the same as the number of figures in its square root. The square of
the highest denomination of the root would be found in the left hand
period ; the square of the two highest denominations would be found in
Uie two left hand periods ; &c.
1. Separate 8478695 into periods, and explain their uses.
Annoer. 8478695. The left hand period, 8, contains all of the square
of the thousands of the root ; the two left hand periods, 847, the square
of the thousands and hundreds ; &c.
Separate each of the following numbers into periods, and
explain their uses : —
2. 5794865. I 4. 375486792.
3. 89475948. 5. 32500075.
J81<1. Method of forming a Squart.
{a ) To find a law of universal application in squaring or extracting
ibe square roots of numbers, we will use the letter a to represent any
anmber whatever and b to represent any other number.
28*
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190 METHOD or FORXIKG A 8QUARB.
(*.) Then will a + b fepresent the sam, and (a + b)« or (a -f* ^
X (a -f- b) the squnre of the sum of any two nnmbers whatever.
(c) Performing the mnltiplication, we hare a times a = a*; a timef
b » a X b, or, as it mnv be written, ab; b times a = a times b =r- a
X b, or ab ; b times b a b>.
(d.) Writing the work, ti below, and adding the partial prodocts, w
bare,—
a+b
ax (a + b) = a« + a times b n= a« + ab
bX(a + b)= +atime8 b +bJ=» ab + b«
(a + b) X (a + b)=a» + 2timesab + b« = a? + 2 ab + b*
(e.) Hence, (a + b)s «= a* 4* 2 ab + H oi*! «"<» a« eqnals the
■qnare of the first number, and 2 ab eqnals twice the product of the
first number by the second, and b^ eqnals the square of the second ;
The square of the sttm of any ttco numbers equals the square of the Jirst,
plus twice the product of the first by the second^ plus the square of the
tecottd.
Illustrations.
(7 + 5)» = 75 + 2 X 7 X 5 + 52 =r 49 + 70 + 25 = 144 = 12«
(8 4- 4)« = 8« 4- 2 X 8 X 4 4- 42 = 64 + 64 + 16 = 144 = 128
(20 + 3)« = 202 + 2X 3X20 + 33 = 400 + 120 + 9 = 529 = 23«
(/) But a* + 2 ab + b2 can be put in|o another form ; for 2 ab +
bt = 2 a times b + b times b, = (2 a + b) times b, or (2 a + b) X b.
or, by omitting the sign X» as may be done without ambiguity, (2 a +
b)b.
Hence, (a + b)« = a« + 2 ab + V = aS + (2 a + b) b.
(g.) But a2 means the square of the first number ; 2 a + b means the
tnm of twice the first number, plus the second ; and (2 a + b) b the
fum of twice the first, plus the second, multiplied by the second.
(h.) Jlencey the square of the sum of two numbers is also equal to the
equare of ihe first number^ plus the product obtained by multiplying the sum
tf twice ihe first number plus the second^ by the second.
Illustrations,
(7 + 5)2 = 72 + (14 + 5) 5 = 49 + 95 = 144 = 12«.
(8 + 4)2 = 82 + (16 + 4) 4 = 64 + 80 = 144 = 122.
(40 + 8)2 = 402 + (80 + 8) 8 = 1600 + 704 = 2.304 = 481.
(t.) Now, as any number above ten is composed of tens and units,
't» square will be composed of the square of the tens, plus the prodaci
»f twice the tens plus the units multiplied by the units.
ij.) If there are more than ten tens in the number, the part which if
\
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METHOD OF BXTRACTING THE SQtJARE ROOT. 8Sl
composed of tens may be considered as made np of hundreds and tens,
and its square will equal the square of the hundreds, plus the product
of twice the hundreds, plus the tens, multiplied by the tens.
(k.) Proceeding in this way, we shall at last reach the part which is
expressed by one or two figures, and composed of only the two highest
denominations of the given number. The square of this part will be
the square of the highest denomination, plus the product of twice the
highest denomination, plus the next lower, multiplied by the next lower
Thus, —
(4837)2 =- (4830 + 7)2 «= 48302 + (2 X 4830 + 7 ) X 7
(4830)2 = (4800 4- 30)2 = 43002 + (2 X 4800 + 30) X 30
(4800)2 — (4000 + 800)2 =- 40002 + (2 X 4000 + 800) X 800
21S» Method of extracting the Square Root
Wliat is the value of V 925444 ?
Solution, — (a.) Since this number lies between 10,000 and 1,000,000,
its root must lie between 100 and 1000, and must therefore be composed
of hundreds, tens, and units. Dividing it into periods of two figures
each, it will take the form 925444.
(6.) If, now, we let a represent the hundreds of the root, and b the
tens, the whole of a2 will be found in the left hand period, L e., in the
ten-thousands, and the whole of (a + b)2 in the two left band periods,
L e., in 9254 hundreds.
(c.) The greatest square in 92 is 81, the root of which is 9. There-
fore, 9 = a = the hundreds figure of the root. Subtracting a^, = 81
ten-thousands, from 92 ten-thousands, leaves 1 1 ten- thousands, to which
adding the 54 hundreds gives 1154, which must contain (2 a + b) b.
(d.) Now, as we know a, we can find 2 a, and make use of it as a
trial divisor to find b. But a being hundreds and b tens, 2 ab must be
thousands, and no part of it will be found to the right of the thousands.
(e.) Hence, in dividing, we may disregard the right hand figure of 11 54,
and see how many tinges the trial divisor, 18, is contained in 11.5. The
quotient is 6, which is probably b, the tens figure of the root. If this is
correct, (2 a + b), or the true divisor^ must be equal to 186, and (2 a -f-
b) b must be equal to 6 times 186, or 1116. This last product, being
less than 1154, shows that the work is correct. We subtract, and to the
remainder, 38 hundreds, add the right hand period, 44 units, which gives
8844 for a new dividend.
(/) Now, if we let a' represent the part of the root already found,
i. e., the 96 tens, and b' the units, a' -f b' will represent the requhred
root, and (a' -f b')2 = a'2 _|- (2 a' -f b') b' the given number.
(g.) But we have already subtracted a'2 ; the remainder, 8844, 1
tontain the (2 ;i + ^') ^'*
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882 suLB roR squark root.
(L) We now mike S a', or 192, a trial divisor; and nnoe 3 a'V
most be tent, we omit the right hand figure of 3844, and see how many
timet S a', or 192, it contained in 384.
{L) Thit giret b' >■ 2, >■ the probable nnits figure. If it be correct^
the tme divisor, 2 a' + b^^matt equal 1922, and (2 a' + b') V must be
S X 1222. Thit, being equal to 3844, shows that the given number is a
perfect tqnare, and 962 it its root.
(j,) If there had been another figure in the root, we might have
represented the part of the root already found by a", or by some othef
letter, and the next figure by b", or by some other letter, and have pro*
ceeded as before.
(k,) The numerical work wotild be written thus : —
925444 ( 962 =» Boot
8l = a»
Sa + b«Kl86 ) 1154
1116s=(2a + b)b
S a' + V » 1922 ) 3844
3844 == ( 2 a' + V ) V
SIIO* Rule for Square Root, vnth ProUems.
As a similar process can always be followed, we -may describe the
method of extracting the square root of a number thus : —
First Divide Uie given number into periods of two fgwres^ beginning
with the units.
Second. Find the greatest square in the left hand period^ and place tta
root as the highest denomination of the required root.
Third. Subtract the square thus found from the left hand period^ and to
the remainder bring down the next period^ calling the result a dividend.
Fourth. Double the part of the root already found for a trial divisor.
Fifth. See how many times this trial divisor is contained in all of the
dividend^ excepting the right hand figure^ and write the quotient as the next
figure of the rooty and also place it at the right of the trial divisor^ to form
a true divisor.
Sixth. MukipHy this true divisor by the root figure last founds and mi^
ka:t the product from the dividend.
Seventh. Bring down the next period to the right of the remainder, U
firm the next dividend.
Eighth. Double the part of the root already found for a trial divisor, and
proceed u indicated in the 5<A, 6M, 7M, and Sdi of these paragraphs.
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8QDARB BOjOT OF FBACTIONS. 9ht
What is the square root of each of the following num-
bers: —
1. 7056.
2. 9025.
8. 3104644.
4. 349281.
5. 4137156.
6. 22610025.
7. 4260096.
8. 38580769440964.
317. Square Hoot of Fractions,
(a.) Since the square of a fraction equals the square of its numera-
tor divided by the square of its denominator, the square root of a frac>
tion must equal the square root of its numerator divided by the square
root of its denominator.
'^^'^^ J\ = \ ^""^ (§) = \ '^'^^ '^' ^""'^ ^•^'' ^ '^^'
(6.) If the numerator and denominator of a fraction are not perfect
squares, we can only get the approximate value of its square root
(c.) In such cases, if the denominator is not a perfect square, it wfl'
be well to multiply both terms by such number as will make it so. Thif
number may be either the denominator, or the product of the prime nnm«
bers which are found as factors in the denominator, 1, 3, 5, or any odd
number of times.
15
'36
Thus,
7
19
7 X 11
11 X 11
19
_ 77 5
121' 12
19X2X3
5
22 X 3
114
5
X3
22
490
10000
X 32
24
.049
2«X 3
=s .0490, 01
24 X 32 144
49 49 X 10
1000 1000 X 10
79
{d.) What is the square root of ^ ?
-, . ^9_ 79 _ 79 X 3 X 7 _ 1659
iioetaton. ^ 22 X 3 X 7 22 X 32 X 72 22 X 3^ X 7«
„ /79" / 1659 __ >/ 1659 40 .
"^°^' V 84~ V 22X32X72- 2X3X7 =" 42' ^PP""**'
mately.
This differs from the true root by less than ^V*
(e.) Should a greater degree of accuracy be required, both terms
may be multiplied by such a square number as will make the denomina*
tor sufficiently large to secure the requisite degree of accuracy.
Thus, multiplying the numerator and denominator of rj ^ ^
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884 BQUAJtB BOOT OF DECIMAL FRACriONS.
by 9«, KiTM ^ ^ ^ ^x'l*~X~i' *^.® *PP"**"°*^® "^* ^^ ^^<* *
SOS 203
■ - . ^ , >= o,^i which differs from the true root by less this
1
SIO*
What 18 the square root of each of the following frao
tions, —
1. /AV? I 8- *? I ^- i^
2. igJI? !• 4. I? I 6. tV?
918* Square Root of Decimal Fractions.
(a.) In order that a decimal fraction may have a perfect sqnare for
Its denominator, it must hare an even number of places in its numer-
ator.
Thus, the denominators of .04, .25, .17, .6561, and .384736 are per-
fect squares, but the denominators of .4, 2.5, .017, .06561, .0384736, &C.,
•re not
(6.) If a decimal fraction, the root of which is required, does not
have an even number of decimal places in its numerator, a zero must
be annexed, to make the number even, so that the denominator in all
cafrM may be a perfect square..
(c.) Since ^"!oi = /— == — = .1, v/ioooi = /— — =«-—
^' ^ VIOO 10 I V-w*^* V 10000 100
■■ .01, Ac, it follows that there will be as many decimal places in the frac
tional part of the root as there are times two decimal places in the frac-
tional part of the power. Hence, we can carry out the root to as many
places as we choose, by annexing two zeros to the power for every
additional figure which we wish to obtain in the root
Required ^J A
Solution.^ Since ^/A = ^/AO = y.4000 == y .400000, we may hare
the following written work : —
.40 ( .6324
.36
1.23 ) 400
369
1.262 ) 3100
2524
1.2644 ) 57600
50576
7024
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CUBE ROOT. — RELATION OF CUBE TO ROOT. 385
The exact root of such a number as the above cannot be found, fof
A number which does not end with a zero cannot have zero for thd
right hand figure of its square.
What is the square root of —
1. .0625?
2. 5.625?
3. .8281? I 5. 56.25?
4 1.6? I 6. .16?
319. Cube Root. — Relation ef Cube to Root
(a.) To find a method of extracting the cube root of a number, wa
must make some preliminary investigations similar to those of dl8
and 314*
TA.BLB OF BOOTS AND CUBES.
1» = 1 48 =r 64 7« = 343
2» = 8 58=125 88 = 512
3* = 27 6' = 216 98 = 729
108 -- 1,000 1,0008 = 1,000,000,000
1008 =s 1,000,000 10,0008 = 1,000,000,000,000
(6.) The above table shows that, —
First There are below 1000 but 9 entire numbers which are perfect
cubes.
Second. The entire part of the cube root of any number below 1000
will be less than 10, and therefore will contain but one figure ; of any
number between 1000 and 1,000,000 will lie between 10 and 100, and
therefore will contain but two figures ; &c.
How many figures are there in the entire part of the root
of 47986754?
Answer. — Since 47986754 lies between 1,000,000 and 1,000,000,000,
its root must lie between the roots of those numbers, L e., between 100
ftnd 1000, and hence must contain 3 figures in its entire part
330* Division into Periods.
(a.) As the cube of 10 is 1000, of 100 is 1,000,000, &c., it follows
chat the cube of any number of tens will be some number of thousands,
of any number of hundreds will be some number of millions, &c. ; or,
in other words, that the cube of tens will give units of no denomination
below thousands, the cube of hundreds will give units of no denominsii
tion below millions, Ac.
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•M MXTHOD or FORxnro ▲ cubb.
(I.) HflB«e, the three right hand figures of any nvmber will contain
so pari of the cobe of the denominationi abore units, the six right hjukd
figures will contain no part of the cube of those above tens, &c.
(e.) Therefore, if we should begin at the right of any number, and
•eparate It into periods of three figures each, the number of periods
would be the same as the number of figures in its cube root The cabe
of the highest denomination would be found in the left hand period, the
oibe of the two highest denominations would be found in the two left
hand periods, &c
1. Separate 9876585925 into periods, and explain their
rues.
AnMwtr, 9870585925. The left hand period, 9, contains all of the
eabe of the thousands of the root ; the two left hand periods, 9876, the
cobe of the thousand and hundreds ; &c
Separate each of the following numbers into periods, and
explain their uses : —
2. 42783794 I 4 5847643759427.
8. 584376423. | 5. 6972842903612.
331 • Method of forming a Ckihe,
(a.) To find a law of universal application, we will use die letter a
to represent any number whatever, and b to represent any other number
(6.) Then will a + ^ represent the sum, and (a + b)*, or (a + b)
X (a + b) X (a + b), the cube of any numbers whatever.
(c.) But (a + b)»r= (a + b)* X (a + b) = (a« + 2 ah + b*) X
(a + b). Now, multiplying a2 4-2ab + b2bya gives a* + 2 a^ b +
ab*, and multiplying it by b gives a^ b + 2 ab> + b^. Adding these
results together gives a' -f- 3 as b -f 3 abs -f bs.
(d,) This work may be written out thus : —
ali + 2ab + b8=(a + b)«
t + b
a? + 2a?b + ab» = a X (a» + 2 ab + b^) .
a«b4-2ab»-f-b8 = bX (a»4-2ab + bs)
a^-f-3a^b + 3abs + b3=(a + b) X (a»4-2ab4-b2)=:(a+b)».
(s.) Bat3a?b + 3abs + b8 = 3as times b + 3 ab times b + hs
tim«sb««(3 aJ + 8 ab-f-bz) times b, = (3 a8 + 3 ab + bs) b.
Again : 3 ab -f b^ s= 3 a times b -f b times b »: (3 a -f b) timef
a » (3 « + b) b.
Hence, a« 4. 3 a* b -|-.3 ab^ + bs = a8 4. [3 aa -K{8 a + b) bj X b;
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TO EXTRACT THE CtJlil ROOT. 587
Le^ihe cube of every number composed of two parts is equal to the cube of
the first part; plus the product obtained by multiplying the second part by
the sum of three times the square of the first part^ plus the sum of thrm
times the first part plus the second multiplied by the second.
888. To extract the Cube Root
Wlmt ia the cube root of 259694072 ?
Solution. — (a.) Since the namber lies between 1,000,000 and 1,000,-
000,000, its root most lie between 100 and 1000, and hence must contain
Ihrce figures. Separating it into periods of three figures each, it uriU
take the form 25969407*2.
(6.) If we now let a represent the hundreds of the root and b the
tens, it is evident that the whole of a^ will be found in the left hand
period, and of (a + h)^ in the two left hand periods.
(c.) The greatest cube in 259 is 216, the root of which is 6. There-
fore, 6 -= a, sss the hundreds figure of the root. Subtracting a^, = 216
millions, from 259 millions leaves 43 millions, to which adding the next
period gives 43694 thousands. We may regard this as a dividend, and
it must contain [3 a* -f- (3 & + h) X h] b, i. e., the remaining part of
(a + b)8.
(d.) Now, as we know a, we can find 3 a^, and make use of it as a .
trial divisor to find b. But a being 6 hundreds, 3 a^ must equal 108
ten-thousands, and as b is tens, 3 a b must be hundred-thousands.
Hence, we may disregard the two right hand figures of the dividend,
and see how many times the trial divisor, 108, is contained in 436.
(e.) The quotient being 4, we write it as the probable tens figure of
the root, and have next to complete the true divisor, 3 a^ -f* (3<^ + b) b.
But (3 a 4- b) = 18 hundreds + 4 tens = 184 tens, and (3 a + b) b
Bs 184 tens multiplied by 4 tens = 736 hundreds. Hence, the true
divisor 3 a» + (3 a -f- b) b = 108 ten-thousands + 736 hundreds =a
11536 hundreds.
(/) Multiplying this by 4 gives [3 a^ + (3 a + b) b] b = 46144
thousands, which, being greater than 43694 thousands, shows that then
are not as many as 4 tens in the root, and that b is less than 4.
(g.) Assuming b = 3, and proceeding as before, we find that the
true divisor 3 a' + (3 * + h) b = 108 ten thousands -|- 549 hundreds
B= 11349 hundreds; and [3 a> + (3 a + b) b] b = 34047 thousands,
which, being less than 43694 thousands, shows that 3 is the true tens
figure of the root Subtracting 34047 thousands leaves 964fthousand8,
to which adding the next period, 072 units, gives 9647072 for a new
dividend, which must contain the remaining part of the power.
(A.) If we now let a' represent the part of the root already found,
29
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us TO EXTRACT THE CUBE ROOT.
L e^ tka €3 tent, and b the units, we shall hare S5969407S = a" -^
I« a « + (8 a' + V) V] V, and, as we hare already subtracted a«
•647072 wfll contain [3 a^ + (3 a' + V) b'j V.
(1.) Bat 8 a'* MB 3 times the square of 63 tens == 11907 hundreds^
which maj, as before, be made use of as a trial divisor to find V.
As 3 a** is hundreds and V is units, 3 a'' b' must be hundreds ; hence,
•o part of 3 a'' h* can be found to the right of hundreds, and wo maj
disregard the two right hand figures of the dividend, and see huw manj
limes the trial divisor, 11907, is contained in 96470.
(/.) The quotient behag 8, we write 8 as the probable units figure of
tiie root, and complete the true divisor, 8 a* -f* (3 * + h) b. 3 a -f- 1
iB 189 tens + 8 units « 1898 units, and (3 a + b) b = 1898 X 8 =1
15184. Therefore, 3 a< + (3 a + b) b = 11907 hundreds + 15184
vnhs H 1S05884 B the true divisor.
(k.) Mnltipljing this hj 8 gives 9647072, which shows that 8 is the
troe value of b', and 638 is the root required.
Proof. — See if 638« =» 259694072.
(/.) Had the root contained another figure, we might have taken a''
lo represent the part alreadj found, and b" to represent the next fignivt,
when we should have, (a" + b")« = •"» + [3 a"« + (3 a'' + V) b"]
V equal the number.
(la.) Much of the labor of finding the trial divisor, 3 a's might havo
been avoided. For as a' ss a -f* b, 3 a'* must equal 3 times the square
of t + b, or 3 times (al + 2 ab + V) = 3 a« + 6 ab + 3 b«.
(n.) But the previous trial divisor, 3 a« + (3 a + b) b = 3 a« +
3 ab -f~ b*, and the number which stands above it equals (3 a -{- b) b
■■ 3 ab + V ; hence the sum of these equals 3 a' + 6ab + 2b^
which only lacks b* of being equal to 3 a' + 6 ab + 3 b*, or to 3 a''.
Hence, by squaring b, and adding it to this result, we have 3 a' ss 3 a^
+ 6ab + 3bs.
(0.) The work may be written thus : —
. . . SiliV
259694072 (638
216000000 630 s a'
8 a^» 1080000) 43694000
(8a + b)b»i88 X3b 54900*]
[3 a« + (3 a + b) b] = 1134900 f 34047000 = [3 a« +
b««=i 900J (3a + b)blb
3 a' KB 1190700) 9647072
(3 a' + V) b' b' = 1898 X 8 = 15184
f 5 a« 4 (3 a' + V) b'J = 1205884 9647072 = [3 a« +
(3a' + b')b|b
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BULE FOB THE CUBE ROOT. d39
(p.) By keeping in mind the denominations, so as to render the ccrof
unnecessary, we should have the following form : —
259694072 ( 638
216 = a»
Trial divisor == 3 a« s=3 108 ) 43694 = Divid.
(3 a + b) b = 183 X 3 = 5491
True divisor, 3 a« + (3 a -f- b) b, = 11349 f 84047 = [3 a« +
V = 9 J (3 a + b) b] b
Trial divisor == 3 a'2 = 11907 ) 9647072 = Divid.
( a' -f. V) b' = 15184
True div. = 3 a" + (3 a' + b') b' =: 1205884 9647072 = (3 a" +
(3 a' + b') W] y
333. Eule for the Cube Boot
As a similar process will always apply, we may describe the method
of extracting the cube root of a number thus ; —
First. Divide the given number into periods of three figures each^ begin'
ning with tfie units.
Second. Find the greatest cube in the left hand period, and place its rod
€U the first figure of the required root.
Third. Subtract this cube from the left hand period^ and to the remainder
bring down the next period^ calling the result a dividend.
Fourth. Find three times the square of the part of the root alreadjf
founds and make it a trial divisor.
Fifth. See how many times the trial divisor is contained in the dividend,
excepting the tuio right hand figures^ and unite the quotient as tlie nextfigurt
of the root.
Sixth. To three times the part of the root previously founds annex thd
hst root figure^ multiply the result by the last figure, and placing the prodtict
under the trial divisor^ two places to the right, add it to the trial divisor.
This will give the true divisor.
Seventh. Multiply the true divisor by the last root figure, placing the
product under the dividend.
Fightk. Subtract the product from the dividend, and to the remainder
mnnex the next period for a new dividend.
Ninth. Add tlie square of the last quotient figure to the last true divisor
and the number standing over it. The sum will equal three times the square
of the lOot already found, and will be the second trial divisor.
Tenth. Now proceed as directed from the fifth forward.
What is the cuhe root of ea^h of the following niimhei's?-^
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S40
CITBE BOOT OF FEACIION3.
I.
830584
6.
432081216.
2.
262144.
6.
27054036008.
S.
676836152.
7.
30225545875.
4.
188250432.
8.
6804992375.
334. Cube Root of Fractions.
(a.) The cube root of a fraction equals the cube root of its nomcra*
lor, divided bjr the cube root of its denominator ; and if both terms of
the fraction are not perfect cnbes, onlj its approximate root can be
obtained.
(6 ) If in snch cases the denominator is not a perfect cube, it will bo
well to multiply both terftj by the sqnare of the denominator, or by
snch other nnmber as will make it so. Thus, —
8 A"^ 8 /2 X 3« _ 8 A8^^/l
V3^V3X3» V27 I
/3 X 2
X 2'
8 2
]. What is the cube root of ^ ?
6 5 5 X 32 45
Solution, — =
Hence. V — == / —
28 X 38 216*
45 y45 3 . ;,
— = ^ = g, approximately.
This differs from the true root by less than ^.
(c.) If a greater degree of accaracy is required, both teims may ht
multiplied by some perfect cube before extracting the root. Thus, —
y-:
45 X 28
8
V^360
ZisX^-V 1728^"-T2-~l2'^P^"'^*^^^'
which differs from the tme root by less than ■^.
What is the cube root of —
3. A? I 5. if? 1 7.
f?
ftitS. Cube Root of Decimal Fractions.
(a.) In order that a decimal fraction may have a perfect cube for iti
denominator, it must contain 3, 6, 9, cr some multiple of three places in
Its numerator
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RULE FOB EXTRACTING A ROOT OP ANf DEGREE. 341
Thns, the denominators of .008, .027, .512, .003, and .375067 are each
perfect cubes, while the denominators of .8, .08, .0027, and .56789 are not.
(c.) If a decimal fraction, the root of which is required, does not con-
tain 3, 6, 9, or some exact multiple of 3 decimal places, zeros must be
annexed, so that the denominator may be in all cases a perfect cube.
(d.) Since ^iooi = V-A- = -L = .1, and y .000001 = */— !
^ ' ^ V 100 10 ' ^ V 1000000
«» —— ^ .01, &C., it follows that there will be as many decimal places
in the fractional part of the root, as there are times three decimal places
in the fractional part of the power. Hence, we carry out the root to
as many decimal places as we choose, by annexing three zeros to the
power for each additional figure we wish to obtain in the root
What is the cube root to four places of decimals of —
1. .87. I 3. 1.76. I 5. .427.
2. .6735. 4. 29.78. 6. .0007.
330* Jiule for extrcxting a Root of any degree.
A root of any degree may be found as follows : —
First. Divide the given number into periods of as many figures eaeh tm
there are units in the index of the required root.'
Second. Find the greatest power of the required degree in the left hantt
periodf and place its root as the first figure of the required root.
Third. Subtract the poroer from the left liand figure^ and to the remain'
der bring down the first figure of the next period for a dividend.
Fourth. Eaise the part of the root already found to a power one degree
less than the given power^ and multiply the result by the index of the required
root^ calling the result a trial divisor.
Fifthr Divide the dividend by the trial divisor^ and the quotient will prob-
ably be the next figure of the root. To ascertain vheiher it t«, place it in
the root, and raise tfie number thus found to the required power. If the
result equals the first two periods of the given number, or is less, the root
figure is correct ; hut if as will often be the case, it is greater, the root figure
is too large.
Sixth. Having found the true root figure, fina the remanuUr, amdfiirm
m trial ditfisor, ^ as before.
29*
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Mi
mNSURATIOW.
SECTION XVII.
MENSURATION.
337* Polygons.
(c.) k TRiAHOLB IS A fig:ure having three sides and three angles.
(6.) A siOHT-ANOLBD TRIAMOLB IS a triangle having a right anglcu
(c.) For definitions of the angle, RSCTA2I0I.B, jlnd squasb, see
pages 33 and 34.
(d.) Lines arc parallel when they lie in the same direction \ as, for
ioitance, the lines forming the sign of equalitj.
(e.) A pARALLELOOBAic is a foor-sided figure, having its opposite
sides parallel.
(/) A TRAPEZOID is a fonr-sided figure having two of its sides
parallel
(^.) A POLTOOX is a figore bonnded on all sides by straight lines.
Figl. Fig 2, Flg.a,
B C
X
A D
-Q, Y M V N
(A.) Fignres 1 and 2 represent triangles ; figure 2 represents a right-
angled triangle ; figure 3 a square ; figure 4 a parallelogram ; figure 5 a
trapezoid ; and all represent polygons.
(i.) Similar figures are those which have the same shape, i. e., which
have the angles of the one equal to the corresponding angles of the
other, and the sides about the equal angles proportional.
(/) The BASE of a figure is the side on which it is supposed to stand*
Any side of a figure may be regarded as its base. The side opposite the
base of a rectangle or parallelogram is often called its upper base.
Thus, in figure 3, A P is the lower, and 3 C the upper buse-
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MENSURATION. 343
{k.) The altitude of a triaifgle is the perpendicular distance from the
side assumed as its base to the vei-tex of the opposite angle
Thus, in figure 1, when M JST is taken as the base, the distance O R
is the altitude ; in figure 2 when A B is the base, A C is the altitude ;
when B C is the base, A D is its altitude ; and when A C is the base,
A B is the altitude.
(L) Th3 ALTITUDE of a rectangle, a parallel ogram, or a trapezoid is
the perpendicular distance between its parallel bases.
Thus, A B is the altitude of figure 3, z q is the altitude of figure 4,
and u V of figure 5.
{m.) A DIAGONAL of a polygon is a line drawn from the vertex of
two angles lying opposite to each other.
Thus, B D and B C are diagonals of figure 3, z y and x p are diag-
onals of figure 4, F K, F J, G E, &c., are diagonals of figure 6.
(n.) The area of a square or of a rectangle equals its length multi*
plied by its breadth. — See 40, (c), Note, and 163, Note.
(o.) The area of a triangle equals half the product of its base by its
altitude.
(p.) The area of a parallelogram equals the product of its base by
its perpendicular height.
{q.) The area of a trapezoid equals half the product of its altitude
by the sum of its parallel bases.
(r.) The area of an irregular polygon can be found by dividing it
into triangles.
(s.) The areas of diff*erent triangles, squares, and parallelograms are
to each other as the product of their bases by their altitudes.
{t.) The areas of similar polygons are to each other as the squares
of their like dimensions.
(m.) The circumference of a circle equals very nearly 3.1416* times
its diameter.
{v.) The circumferences of circles are to each other as their diame
ters or radii.
(w.) The area of a circle equals half the product of its circumference
by its radius, or quarter the product of its circumference by its diameter.
( r.) The area of a circle also equals the square of its radius muItl' •
plied by 3.1416.
(y.) The areas of circles are to each other as the squares of their
diameters or radii.
♦ More accurately, 3.141592653589 ; but the above is sufficiently exact
for most purposes. Indeed, 3'f is sometimes used where no great degree
of accuracji s required.
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944 MCNSURATIOir.
938. ProhUnu.
General Direction. — Draw figures to coirespond |l
Ihe conditions of each problem.
1. What is the area of a triangle of which the base is 8 &
and the altitude 6 ft ?
2. What is the area of a parallelogram of which the bas«
i« 9 A. and the altitude 6 ft ?
8. What is the area of a trapezoid of which the upper
tmse is 6 ft, the lower 10 ft, and the altitude 9 fl. ?
4. What is the circumference of a circle of which the
diameter is 16 ft ?
5. What is the diameter of a circle of which the circum-
ference is 25 ft ?
6. What is the area of a circle of which the radius is 4 ft ?
7. What is the area of a circle of which the circumference
is 22 ft?
8. What is the radius of a circle of which the area Is
627 ft. ?
9. How many rods long is the side of a field which con-
tains 40 acres ?
10. How many rods in diameter is a circular field which
contains 8 acreo 't
11. What must be the length of the side of a square field
which is equal in area to a circular field 100 rods in circum-
ference ?
12. A rectangular field containing 100 square rods is twice
as long as it is wide. What is its length ?
13. What must be the diameter of a circular field which
contains 4 times as much surface as a similar field 32 rods m
diameter ?
14. The diameter of one circular field is twice that of an
othsr How do their areas compare ?
15. How many square feet in a '>oard 15 feet long, 2 feet
wide at one end, and 1 foot wide a the other, allowing that
the sides taper regularly ?
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PBOPEBTIES OP THE RIGHT-ANGLED TRIANGLE. 345
929* Properties of the Right-angled Triangle^ with Proh*
lems.
Square op the Hypothenuse.
(a.) The side opposite the right angle of a right-angled triangle U
called the hypothenuse.
In figure 2 the side A C is the hypothenuse.
(6.) The square of the hypothenuse equals
the sum of the squares of the other two
I Jes.
(f.) The annexed figure will illustrate the
lutuning of this. Its truth can be rigidly
ttcmonstrated by geometry.
ABC represents a right-angled triangle
ngnt- angled at B. Let A B be 4 ft, and B C
3 ft. Then will A C be 6 ft, and A B^-f B C*
-=AC', orl6-f 9 = 25.
1. What is the hypothenuse of a right-angled tiiangle of
which one side is 6 ft. and the other 11 ft. ?
Svggestion. — The hypothenuse equals ^36 + 121 = ^57 ft
2. What is the third side of a right-angled triangle of
^'hich the hypothenuse is 12ft. and the given side 9 ft^ ?
Suggestion. ^144 — 81 = .^63"= Ana.
3. What is the distance from one corner of a floor to the
opposite corner, if the floor is 24 ft long and 18 ft. wide ?
4. What is the diagonal of a square 30 ft. on a side ?
5. A boy, flying his kite, found that he had let out 845
yards of string, and that the distance from where he stood to
a point directly under the kite was 676 yards. How high
was the kite ?
6. A certain window is 20 ft;, from the ground. How long
must a ladder be which, having its foot 15 ft^ from the bottom
of the building, will just reach the window ?
7. How long is a side of the greatest square whicJi can be
inscribed in a circle 3 feet in diameter ?
Note. — The diagonals of a square bisect each othAj: at right angles.
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846 SOLIDS.
8. A tower, 60 feet high, stands on a mound 30 feet abo\'e
a horizontal pknc. On this plane, and directly south of tlit
tower, is a spring, and directly east from the spring, and at a
distance of 160 feet from it, on the same plane, stands a lai*ge
oak tree. Now, allowing that the distance in a direct line
from the top of the tower to the spring is 150 feet, what is
the distance from the top of the tower to the foot of the oak ?
9. Two men started from the same place, and travelled, one
north at the rate of 4 miles per hour, and the other east at
the rate of 3 miles per hour. After travelling 7 hours, they
turned, and travelled directly towards each other at the same
rate as before, till they met How many miles did each
travel?
10. There is a rectangular field 100 rods long and 80 rods
wide, the sides of which run north and south. A man started
from the south-west comer, and travelled due north along the
western boundary of the field for 60 rods, when he travelled
across the field in a straight line to the north-east corner.
How much farther did he travel than he would if he had
gone in a straight line all the way ?
330. Solids.
(a.) A SFHEBB is a solid bounded by a curved surface, every part of
which is equally distant from a point within, called the centre.
(6.) A line drawn from the centre to the surface is called a badius,
and a line drawn from any point in the surface through the centre to thb
opposite point is called a diambibb.
(c.) A PBI8M is a solid having its several faces parallelograms, and
its bases two equal and parallel polygons.
(d.) A CUBB (see 41* 6.) is a kind of prism.
(«.) A OTLiNDEB is such a solid as would be formed by revolving a
rectangle about one of its sides. It has also been defined to be ** a
round body with circular ends."
(/) A PTBAM ID is a solid body bounded laterally by triangles, of
whi( h the vertices meet at a common point, and the bases terminate in
the sides of a polygon, which forms the base of the pyramid.
{g.) A CONE is a solid which 'as a circular base, and tapers regularly
to a pyini called the vertex.
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SOLIDS.
847
(h.) A iRUSTi/M of a cone or pyramid is a part cut off by a plane
parallel to the plane of its base.
(t.) Similar solids have the same shape, i. e., the angles of one of
them equal the corresponding angles of the other, and the sides about
the equal angles are proportional.
All spheres are similar. Two cones, or two cylinders, are similar
when their altitudes are to each other as the radii or diameters of their
Fiff,l.
Fig- 2.
Fig.S.
fd
Kl
J^
P^
fl^.4.
F^.5.
F^.^
{j\) Figure 1 represents a sphere; figure 2 a prism; figure 8 « cyl-
inder ; figure 4 a pyramid ; figure 5 a cone ; figure 6 a frustum of »
cone.
{h.) The sultFAOB of a sphere equals the square of its diameter
multiplied by 3.1416.*
{I) The surfaces of a\ heres are to each other as this squares of their
radii or diameters.
{m.) The soliditt, or solid contents, of a sphere equals the
product of the surface multiplied by ^ of the radius, or by ^ of the
diameter, or it equals i of the cube of the diameter multiplied by
8.1416.*
(n.) The solidities of spheres are to each other as the cubes of theif
radii or diameters.
* See foot notei page 343.
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348 SOLIDS.
(o.) The solidities of sin. Ur solids are to each other as the cabes of
their like dimensions.
{p.) The soiiditjr of « prism equals the area of its base multiplied by
Its altitade.
(9.) The solidity of a cylinder is eqaal to the area of its base mnki-
plied by iu altitude.
(r ) The convex surface of a cylinder is equal to the circumfereuco
of its base multiplied by its altitude.
(1.) The solidity of a cone or of a pyramid equals the area of itc
base multiplied by i of its altitude.
(< ) The solidity of a frastum of a cone, or of a pyramid, equals |>
of the product of its altitude multiplied by the sum of its upper base, *
plus its lower base, plus the mean proportional between the two bases.
NoTB. — The mean proportional of two numbers is the square roo*
of theur product Thus, the mean proportional of 4 and 9 = ^4 X 9
231. Problems.
1. What is the solidity of a sphere the diameter of which
is 8 feet?
2. What is the surface of a sphere the radius of which is
Ifoot?
3. What is the diameter of a sphere of which the solidity
is 10 feet?
4. What is the circumference of a sphere the solidity of
which is 12 feet?
5. What is the diameter of a sphere of which the surface
is 6 feet ?
6. What is the solidity of a prism of which the altitude is
9 feet, and the base contains 10 square feet ?
7. What is the solidity of a cylinder of which the altitude
is 6 feet and the radius of the base 2 feet ?
8. What is the convex surface of a cylinder of which the
diameter of the base is 5 feet and the altitude 4 feet ?
9. What is the solidity of a cone of which the altitude is 9
feet and the circumference of the base 10 feet ?
10. What must be the diameter of a sphere which contains
8 times as many cubic feet as one 3 feet in diameter ?
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PROGBESSIONS. 349
SECTION XVIII.
PROGRESSIONS.
333* Arithmetical Progression.
(a.) A SEBIES OP NmiBERS IN ARITHMETICAL rROGRRS-
WON, or an arithmetical series, is a series of numbers
each of which differs from the preceding by the same number.
(h.) Such a series would be obtained by continually adding
the same number to, or subtracting it from, any given number.
Thus we should have —
By adding 2*8 to 1, 1, 3, 5, 7, 9, 11, &c
By adding 7*8 to 3, 3, 10, 17, 24, 31, 38, 45, &c.
By subtracting 4's from 29, . . . 25, 21, 17, 13, 9, &c
By subtracting 3*s from 56, . . . 53, 50, 47, 44, 41, &c
(c.) If the series is formed by addition, it is called an
increasing series ; if by subtraction, it is called a de-
creasing SERIES.
(d,) The numbers composing a series are called th6 terms
of the series.
(e.) The difference between the consecutive terms of any series is
called the common difference, and is always the number by the
addition or subtraction of which the series is formed.
(/) Since the terms of a series are formed by continual additions or
subtractions of the same number, it follows that the second term of any
series equals the first, plus cr minus the common difference ; that the
th^rd equals the first, plus or minus twice the common diff^erence ; that
the fourth term equals the £rst, plus or minus three times the common
difference; &c.
(g.) JBencey any term of an arithmetical series is equal to the first term,
plus or minus the common difference taketi one less times than there are temut
in the series ending with the required term,
(h.) Moreover^ if the first term of an increasing arithmetical series be
subtracted from the last^ or if the last term of a decreasing series be sub*
tracted from the first, the remainder will be the product of the common difi
fsrmce multiplied by one lesi than the number of terms,
30
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850 PROGRESSIONS.
233. Problems.
1. Wh&t is tbe 10th term of the increasing sories ol
which the first term is 3 and the common difference 8 ?
2. What is the 25th tena of the decreasing series of which
the first term is 85 and the common difference 2 ?
3. What is the common difference of the series of which
the 1st term is 7 and the 13th term 43 ?
4. How manj terms are there in the series of which the
1st term is 8, the last term 85, and the common difference 7 ?
5. What is the common difference of the series oi which
596 is the 1st term and 491 the 22d ?
6. How many terms are there in the series of which 12 is
the first term, 4 the last, and \ the common difference ?
334. To find the Sum of a Series.
{a.) If we shoald inyert any series, we should have a new one, which
would differ from the former only in the order of its terms, tbe one be*
ing an increasing while the other is a decreasing series. The first term
of one series would equal the last cff the other, and each term of one
Mries would be as much greater than its preceding term as each of the
other 18 less than its preceding term. Hence, if we should write the two
series under each other, and add together the corresponding terms in
the order in which they stand, the successive sums would equal each
other, and each would equal the sum of the first and last terms of the
original series.
(6.) Moreover, there would be as many such sums as there are terms
in the series. Hence, the sum of the two series, or (since they are
equal) twice the sum of either of them, is equal to the product obtained
by multiplying the first and last terms by the number of terms.
Thus, by inverting the series 3, 7, 9, &C., lo 35, we have, —
8 7 11 15 19 'i3 27 31 35 = given series.
35 31 27 23 19 15 II 7 3 =s same series inverted.
38 38 38 38 38 38 38 38 38 = the sums of the sntces-
sive terms.
(<?.) Adding these last results together would give the sum of the
two series, or twice the sum of either, which would manifestly be equal
to 9 times 38, or 9 times the sum of the first and last terms. Dividing
this by 2 would give the sum of one of tlie series.
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GEOMETBICAL PR06KESSI0N. 851
(d.) Hence^ the sum of a series in arithmetical progression equals half
the product obtained by multiplying the sum of the first and last ta-ms by th-
number oj terms,
33«S. Problems.
J. What is the sum of the series of which 9 la tlie Isi
term and^94 the 20th ?
Ansu^. (^ -h 9^) X 20 ^ ^^^ x 10 = 1030.
2. What is the sum of the series of which 427 is the 1st
term and 187 the 81st ?
3. What is the sum of the series of which 4 is the 1st
term and 9 is the 6th ? What is the common difference ?
4. What is the sum of the series of which the 1st term is
7. the common difference 9, and the number of terms 15 ?
5. How many terms are there in a series of which the
sum is 648, the 1st term 3, and the last term 78 ? What is
the common difference ?
6. What is the 1st term and common difference of a
series of which the last term is 164, the number of terms 12,
and the sum 2100 ?
7. Form the series of which the sum is 153, the 1st term
1, and the last term 17 ?
336* Geometrical Progression.
(a.) A series of numbers in geometrical progression, or a
geometrical series, is a series of numbers each of which bears
the same ratio to the one which follows it
(h,) Such a series would be obtained by continually multi-
plying or dividing by the same number.
Tbas, beginning with 2 and multiplying by 3, we should have 2, 6,
18, 54, 162, 486, &c.
By beginning with 3072 and multiplying by j-, we have 8072, 1536,
768, 384, 192, 96, 48, &c
(c.) The numbers comprising the series are called the
TEUMS OF THE SERIES.
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852 OEOMETR10AL PBOGRESSION.
(d,) The ratio of each term to that which follows it if
called the common ratio, and is alwajs the number by
which we multiplied to produce the series.
(e.) If it be an increasing series, the common ratio will eqnal a whole
number or an improper fraction ; but if it be a decreasing series, the
common ratio will equal a proper fraction.
(/.) From the method of forming snch series, it is obyions that the
second term must equal the first multiplied by the common ratio ; that
the third term must equal the first multiplied by the second power of
the common ratio ; &c.
{(f,) llencej any term of a geometrical series must equal the product of the
Jirst term multiplied by the common ratio raised to a power one degree lesa
than the number of the term.
(L) Moreover^ if the last term of a geonetrical series be divided by tha
firsts the quotient wilt be the common differenx raised to a power one degree
less than Ute number of the term,
237. Problems.
1. TVliat is the 7th term of the series of which 125 is the
1st term, and 2 the common ratio ?
2. What is the 5th term of the series of which 1 is tJie
1st term and ^ the common ratio ?
3. What is the 9 th term of the series of which 4096 is the
1st term and i the common ratio ?
4. What is the common ratio of the series of which 1 is
the 1st and 81 the 5th term ?
5. Construct the series of which 1 is the 1st and 6561 ia
the 9th term.
6. Construct a series of 8 terms^ having 1 for the 1st
term and § for the common ratio.
238 • To find the Sum of a Geometrical Series.
(a.) If each term of a geometrical series should be multiplied by the
common ratio, a new series would be formed, of which the first term
would equal the second term of the former series, the second term would
equal the third of the former, &c. ; the last terra but one of the new
series would equal the last of the given series. Hence, the first term of
the original series would have no corresponding term in the derived
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6EOMETBICAL PROGRESSION. 953
feries, and the last term of the derived series would have no corresponds
ing term in the original series.
Thus, by multiplying each term of the series 2, 6, 18, &c., to 1458 bjf
the common ratio, we have — •
2 6 18 54 162 486 1458 = given series.
6 18 54 162 486 1458 4374 = derived series = 3 times
given series.
Again. Multiplying each term of the series 32, 8, 2, -, &c., to — by
Ihe common ratio -, we have —
^2^2--—- -— - = given series.
2 8 32 128 *
® 2 - - — - -— - -— s=s denved senes = - of given
2 8 32 128 512 4 *^
series.
(6.) 17ow, as the given series equals once itself, and the derived series
equals the common ratio times the given series, it follows that the dif>
fercnce between the given and derived series will equal the product of
the given series multiplied by the difference between 1 and the common
ratio.
(c.) But, as has been shown, the difference between tlie series equals
the difi^erence between the first term of the given series and the last
term of the derived series.
{d.) Hencey to find the sum of a geometrical series^ we tnaif multiply the
last term by the common ratio, and divide the difference between the product
and the first term by the difference between 1 and the common ratio,
339. Problems.
1. What is the sum of the series of which 2 is the Ist
term, 1458 the last, and 3 the common ratio ?
Solution. 3 times 1458 = 4374, from which subtracting 2 leaves
4372. Dividing this by 3 — 1, or 2, gives 2186 for the sum of the
series.
2. What is the sum of the series of which 32 is the 1st
term, ^^g- the last, and ^ the common ratio ?
Soltaion, x^ X i = ttjWi which taken from 32 leaves 31-f8ff.
This equals 1 — J-, or J, times the required sum. Hence, the sum of
the series equals Sl^-JJJ -f- i = ^^^ih
3. What is the sum of the series of which ^ is the Isl
term, 486 the last, and 3 the common ratio ?
30 ♦
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854 CmCULATING DECIMALS.
4. What is the sum of a series of 11 terms of which 16 la
the Ist term and j- the common ratio ?
5. What is the sum of a series of 5 terms of which 5 ii
the 1st tej^ and 3125 the hist* term ?
340* hifimte Decreasing Series.
(a.) As in a decreasing series each term is smaller than the preced-
ing, it follows that if the series, be carried far enough, the terms will
become so small that they may be disreg^arded without affecting sensibly
the snm of the series.
(6.) An infinite decreasing scries will always be of this character,
and hence its snm will equal the quotient obtained by dividing the first
term by the difference between 1 and the common ratio.
1. What is the sum of the infinite decreasing series of
which 4 b the 1st term and ^ the common ratio ?
Answer. 4 + (I — i) = 4 -f- ^ = 4|.
What is the sum of the infinite decreasing series —
2. Of which 1 is the 1st term and ^ the conmion ratio ?
8. Of which 3 is the 1st term and f the common ratio ?
4. Of which 2 is the 1st term and | the common ratio ?
5. Of which .37 is the 1st term and .01 the common
ratio ?
6. Of which .597 is the 1st term and .001 the. common
ratio?
7. Of rhich .2794 is the 1st term and .0001 the common
ratio ?
SECTION XIX.
241. CIRCUIATING DECIMALS.
(«.) A circulating or repeating decimal is one which will
never terminate, but in wliich the same figure, or succession
of figures, will always follow each other in the same ordei*.
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CIRCULATING DECIMALS. 35$
Examples. .9999, &c. ; .823232, &c ; • .5174351743, &c. j
.0200602006, &c.
(5.) Circulating decimals are equivalent to vulgar frao-
▼ tions, the exact decimal value of which cannot be found. — -
See 143, (a.)
(e.) That sach vulgar fractions must give rise to repeating dedmali
may be shown thus : —
Since the remainder after any division must be less than the divisor,
we shall at soae stage of the division, explained in 1489 find a remain-
der equal to a former remainder, and from this point the quotients and
remainders will succeed each other in the same order as before.
(d,) A repeating decimal is indicated by placing a dot ovei
the repeating figure, or over the first and last figures of the
repeating period.
Thus, .7 = .7777, &c. ; .19 = .191919, &c. ,
.453142 = .45314231*2, &c.
(e.) The repeating part of a decimal will begin as soon as all the
factors of 10 have been cancelled from the denominator of the vulgar
fraction which produces it, the vulgar fraction being in all cases reduced
to its lowest terms. — Sec 143) (6.)
1. With which place will the repetend of the decimal
value of 2^ begin ?
Answer. — Since 12 contains the third power of 2, which is a factor
of 10, the repetend will commence after 3 places have been obtained,
i. e., with the fourth place.
2. With which place will the repetend of ^ commence ?
Ansroer. — Since 24 contains no factor of 10, the repetend will con^
mence with the first place.
With which place will the repetend of each of the follow^
ing fractions commence ?
3. f. 1 5. ii 7. a.
4- H- 1 6. m. 8. A.
( f.) An expression which contains only the figures of the repetend
is called a single repetend; one which contains other figures it
rallied a MIXED bepetend.
Tb is, .427 is a simple repetend, and .53427 is a mixed repetend.
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S56 OIRCITLATINa DECniALS*
iff,) Two repeteadi tm Bimllar when they begin at the same decimal
place, and oonterminoof when thej end at the same decimal place
Thoa, iS3 and .9, or .S47S, and ^i? are similar; .427 and .436, oi
•4S79, and 037S are contorminoos.
(A.) The repeating period may be considered as beginning at any
igare, provided that it is made to indade the entire combination whi(^
That, il7 ■- 4174 « .41741 « .417417 ; for each developed would
Shre .417417417417, Ac.
(t.) A repeating decimal is really an infinite decr^ising series in
geometrical progression, of which the first term is the first repeating
period, and the common ratio is the decimal fraction, baring 1 for a
•amerator and the power of 10, whose exponent contains as many units
M there are places in the repeating period, for its denominator. (See
hit three pnUetHS in 989.)
Thoa, .48 «: a series of which .48 is the Ist term and .01 the com
non ratio. Hence, .48 = .48 -4- (1 ^ .01 ) « .48 -«- .99 as ||.
Again. 479 ss a series of which .479 is the 1st term and .001 the
aommon ratio. Hence, .479 = 479 -«- (1 — X)01) = 479 -*- .999 «
tta*
So ia98 ■- J298 -f- (1 — .0001 ) = .3298 -4- .9999 m^ HH'
(/.) We should reach the same result by observing that —
i « .111, &c = .1 ; ^ = .010101, Ac. n= .oi ;
vi^ = ^1001, &c. = .OOi ; ^^ir = .00010001, &c. = .0001 ; &c
Now, i — 6X.i=i;.8 = 8X.i = f;
if — 27 X il=tJ; .49 = 49 X .6i=|f5
.147 «=r 147 X .OOi = iH J 045 = 45 X .001 = ^, &C5
4657 « aUI ; .328923 = .32||St.
(it.) Hence, it follows that every repeating decimal is equivalent to a
Tulgar fraction, of which the nnmerator is expressed by the repeating
period, and the denominator by as many 9^s as there are figures in the
repeating period.
What is the value of —
1. .871 I 3. .30061 I 5. .52521
2. .48641 I 4. ^7941 I 6. .2379821
(l.) In multiplying a repeating decimal by any multiple or power of
ten, care must be taken to fill the places left vacant by the change of
the point by the figures of the repeating period, and to observe what
figures w ould have to be added to any period on account of the mnlti^
plication of the preceding period.
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CIRCULATING DECIMALS. 857
Thus, .9 X 1000 = 99.9 .345 X 10 = 8.453
2.743 X 30 = 2.743743, &c., X 30 = 82il2
What is the product of —
1. .4 X 100 ?
2. 27.1436 X 4000 1
3. .84635 X 200000 ?
4. i07 X 100001
5. 28.43 X 1000 1
6. .1374 X 8000 1
(m.) Kepeating decimals, like other fractions, are of the same 4e*
Domination only when they are fractional parts of the same uuiti and
have a common denominator.
(n.) This is the case only when they are similar and conteiminons.
To make circulating decimals similar and conterminous, then, is to
reduce them to the same denomination, which may be done by carrying
them out so far that repeating periods of the same number of figures
beginning and ending at the same place may be formed in each.
1. Make .3587 and .423 similar and conterminous. >
Answer. .35878787 and 42342342, i. e., .35f 5^|J|^ and .42f||Jtf.
Make the repetends in each of the following cases similar and con*
tenninous : —
2. .24 and .375.
3. .467 and .52.
4. .1784 and .328.
5. .5182 and 4.16.
6. .6153 and .4(37.
7. 94287 and 5128i.
(o.) In adding repetends, or in multiplying them, care must be taken
to observe what figures should be added to the written periods, on ac-
count of the addition or multiplication of the next lower periods.
(p.) Thus, in adding .247 + '639 -|- '587, we may observe that since
the sum of the left hand figures is 13, there must be 1 to bring from the
next period below. Adding this with the period gives 1.474.
1. What is the product of .239 X 8 ?
Answer, — Observing that 8 times 39 gives 3 of the next higher de-
nomination, we have 8 X 3.239 = 1.9 15.
What is the value of —
2. 47 + -325 + .i7 + i27?
3. .6279 + .3284568 1
4. .579 X 7 1
6. 69.432746 X 37 1
6. .5183 + 4.6i + 8.85 + .2i7 «
7. 52.17697 + 1.38463 1
8. 29.476 X 12?
9. 43.006 X 8 1
Circulating decimals are multiplied and divided by each other m
Tolgar fractions are.
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a»8
UTSOBLLAMEOUS EZ1MPLE8.
What ii the Talne of —
1.
^79 X A1
5.
42.76 X .437 ?
S.
i79X ^2861
6.
23.84 X 27.96 ?
3.
is -4- .if 1
7.
2.974 H- i007 1
4.
1
5768-4-^1
8.
1.728 ^.0i44f
SECTION XX.
S43* Miscellaneous Examples.
1. How ma j pounds of iron are equal in weight to 100 poands of
gollf
2. 4f times a certain number added to 1 equals f of the quotient ob-
tained by dividing 6f by 1-|. What is the number ?
8. How many seconds are there from 9f o'clock, A. M., of June 4,
1855, to 2i oVIock, P. M., of Aug. 7, 1855 1
4. John can do a piece of work in 5 days, and James can do it in 6
days. How many days will it take both to do it ?
5. How many hogsheads, of 63 gallons each, will a cylindrical cis-
tern, 8 feet in diameter and 10 feet high, hold ?
6. What is the least common multiple and the greatest common
divisor of 74333, 313131, 171717, 146853, and 53095?
5i . 64
7. What is the value of :^ + ^ 1
8. Mr. Taylor has invested -J of liis money in railroad stock, f of it
in bank stock, i of it in real estate, and the rest in trade. Moreover,
what he has invested in railroad stock is $20,000 more than he has in-
vested in trade. How much money has he, and how much has he
invested in each way ?
9. Moses E. Fuller bought 18 shares of bank stock at an advance of 8
per cent on their par value of $100 per share. Six months afterwards,
and at the end of every subsequent 6 months, he received a dividend of
4i per cent. At the end of 2 yr. 3 mo. he sold the stock at a premium
of 12 per cent Money being worth 6 per cent per year, compound in-
terest, how much did he gain by the speculation ?
10. Mr. Goodwin and Mr. Brown commence trade with equal snms of
money. Mr. Goodwin gained $2000, and Mr. Brown lost 10 per cent
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MISCELLANEOUS EXAMPLES.
a5»
of his stock, wten it was foand that Mr. Groodwin had just twice as
mnch as Mr. Brown. What was the original stock of each ?
11. A young man having contracted a debt equal to § of his income,
found that, by saving ^ of his income annually, he could in 5 years
pay up his debt and have $50 left. What was his income ?
12. The hour and minute hands of a watch are together at 12 o'clock
When will they next be together ?
13.' At what time between 12 and 1 will the hour and minute hands
of a watch point in opposite directions ?
14. What was due on the following account, July 1, 1855, interest
being 6 per cent per year?
Dr.
Wm. Barnes, in account with James Sh'edd.
Cr.
1855.
1855.
Jan. 17.
To Sund., 6 mo.
$673
42
Feb. 1.
By Sund., 4 mo.
$237
80
Jan. 28.
To Sund., 4 mo.
542
31
Mar. 5.
By Sund., 3 mo.
492
50
Feb. 13.
To Sund., 6 mo.
237
23
Mar. 21.
By Sund., 6 mo.
873
27
Apr. 22.
To Sund., 3 mo.
720
60
Apr. 25.
By Sund., 3 mo.
594
82
May 10.
To Sund., 2 mo.
54
20
May 27.
By Sund., 4 mo.
376
15
June 23.
To Sund., 3 mo.
133
60
June 2.
By Sund., 2 mo.
142
60
June 20.
By Sund., 6 mo.
225
CO
15. What is tl^ value of a pile of wood 40 ft. long, 4 ft wide, and 5
ft. high, at $5.30 per cord ?
16. A owes B $144, due in 6 mo. 20 da., and B owes A $324, due in
1 yr. 4 mo. and 20 da. If A should pay half of his debt now, and the
other half when, by the conditions, the whole debt was due, when ought
B to pay the whole of his 1
17. A owes B $600 dollars, due in 9 mo., and B owes A $900, due in
15 mo. If A does not pay his debt till B's would otherwise have be-
come due, when ought B, in justice, to pay his debt to A ?
18. A man travelled 100 miles in two days. ^ of the distance he
travelled the first day, added to j- the distance he travelled the second
day, equals j- the distance he travelled the first day. How far did he
travel each day ?
19. A set out from Providence to go to Boston, a distance of 42
miles, and B at the same time left Boston for Providence. At the end
of six hours tbey met, when it appeared that A had travelled 1^ miles
per hour more than B. How far had each travelled ?
20. A dishonest silversmith bought a bar of gold at $192 per lb., and
sold it for $16 per ounce, weighing it in hoth cases by avoirdupois
freight. How much did he gain by the fraud, allowing that the trus
freight of the bar was 5 pounds, and that gold was worth $16 p^i
ponce?
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9G0 MISCELLANEOUS EXAMPLES.
81. A certain room is 16 ft. long, 15 ft. wide, and 12 ft. high. Wliai
if the distance from the right hand upper comer to the left hand lower
comer!
2S. A ressel of war left port with provisions enough to last 600 men
18 months. At the end of three months she captnred and sank an
enemy's ressel, and took on board her crew of 150 men. Two months "
after she captured another vessel, on board which she placed the prison-
ers taken ftom the former prize, and 100 men besides. Four months
after she captured another vessel, on board which she placed 50 men.
When she retumod to port again she had provisions enough to last the
crew which was left on board of her 1 month. How long was she ab-
sent from port t
23. A prize of $31,000 is to be distributed among three officers and
eight sailors, so that each officer shall have 2 j- times as much ai a sailor.
What will be the share of each 1
24. Take any number whatever, multiply it by 3, add 7, subtract the
original number, multiply by 2, add 6, divide by 4, add 27, subtract the
original number, and the result is 32. Why is this ?
25. A merchant sold 50 bushels of wheat for Mr. Randall, and 60
Dushels for Mr. Palmer, receiving $150 for the lot Now, allowing that
Mr. RandalPs was worth 20 per cent more per bushel than Mr. Palmer's,
how ought the money to be divided ?
26. I bought 25,000 feet of boards at $2.25 per thousand, and sold I
of them for what i of them cost What per cent did I gain on the part
soldi
27. I bought 63 kegs of nails, each keg containing 100 pounds, at 4^
cents per pound, and sold $ of them for what i of them cost What
per cent did I lose on the part sold ?
28. I sold I of a lot of land for the cost of f of the lot, and the
remainder for ^ of what I sold the first part for. What per cent of ita
cost did I gain on the entire lot ?
29. A certain sum is to be divided among three persons in such a
way th.U the first has 2 dollars as often as the second has 3 and the
third 6. It turns out that the third has 12 dollars more than both the
others. What was the sum divided, and the share of each person ?
30. A hare starts 30 yards before a greyhound, but is not seen by him
till she has been up 20 seconds. If the hare rans at the rate of 8
miles per hour, and the hound at the rate of 10 miles per hour, how
long will the chase continue, and how far must each ran from his place
of starting ?
31. How many per cent in advance of the cost must a merchant ask
for goods, that, after allowing for a loss of 6 per cent of his sales by bad
debts, an average credit of 6 months, and expenses equal to 8 per cent
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MISCELLANEOUS EXAMPLES. 361
of the cost of the goods, he may make a gain of 10 per ceiU |>f the
original cost, money being worth 6 per cent per year?
32. At a certain time between 12 and 1 the minute hand lacked as
mucli of being at the 1 mark as the hour hand was beyond the 1 2 mark
What time was it?
33. A gentleman bought a horse, chaise, and harness for $450. The
horse cost 3^ times as much as the harness, and Ji£ rhaiso cost $50
more than the horse. What was the cost of each ?
34. George Rice lias 80 yards of broadcloth, which lio will sell for
cash at $4 per yard, but for which he will ask $4.50 per yard in exchange
for other commodities. Edward Tyler has silk for which he charges
$1.25 per yard, cash. How much ought he to charge per yard for it, in
exchange for Rice's broadcloth 1 If, however, he should pay $100 cajh,
and the balance in silk, how many yards of silk ought he to give ?
35. My agent in New Orleans has sold for me a lot of goods for
$3375, on a credit of 6 months, and got the note discounted at a bank.
If he charges 4 per cent for services in selling, and 3 per cent for guar-
antying the note at the bank, how much ought he to remit to me ?
36. July 1, 1851, 1 got my note for $1000, payable in 3 months, dis-
counted at a bank, and immediately invested the money received on it
in land. Oct. 7, 1851, Isold the land at an advance of 12 per cent,
receiving ^ of the sales in cash, and a note fo? the other half, payable
July 1, 1852, without grace, and to be on interest at 7 per cent after Jan.
1, 1852. I lent the cash at 6 per cent interest. When my note at the
bank became due, I renewed it for the same time as before, and at the
proper time renewed it again j and when this last note became due, I
renewed it for such time that the new note would beiome due July 1,
1852. Allowing that I paid 6 per cent interest on the money bon-owed
at the bank, and that I made a complete settlement July 1, 1852, what
was the amount of my gains ?
37. April 16, 1850, I bought of Mr. Curry 498 cords of wood at $3.50
per cord, giving in payment my note payable on demand with interest.
Oct. 5, 1850, 1 sold 232 cords of it at $3.75 per cord, cash, and imme-
diately lent the money received for it on interest. Oct. 17, 1850, 1 sold
the remainder at $4.07 per cord, to be paid Jan. 1, 1851, and to be on
interest after Nov. 1, 1850. Jan. 1, 1851, I collected the money due me,
and paid that due to Mr. Curry. How much did I gain or lose by the
transactions ?
38. How much will it cost to plaster the walls and ceiHng of a room
16 ft. 3' long, 14 ft. 2' wide, and 12 ft. high, allowing for 2 doors, each 7
ft. high and 3 ft. 6' wide, four windows 5 ft. high and 2 ft 8' wide, a fire-
place and mantel 4 ft. 3' high and 5 ft. wide, and a mopboard 6 inchei
wide, provided that it costs 11 crits per yard.
81
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d62 MISCELLxKEOUS EXAMPLES.
SS. An importer sold cloth to a wholesale dealer, end gained 10 per
wnt of what it cost him. The wholesale dealer sold it to a retail detder
at an advance of 10 per cent on what it cost him. The retail dealer
•old it at an adyanco of 20 per cent on what it cost him. Now, allow-
ing iliat the retail dealer received $726 for the clotli, how much did it
cost the importer ?
40. WTiat must be the diameter of a globe which contains 27 times
AS many cubic inches as a globe 2 inches in diameter ?
41. A man spent i and -J of his money, and then earned $36, when
he had SS8 more than -J of what he had at first How much money
had lic at first ?
42. George says that if he should earn as much more money as he
DOW has, ^ as much more, and 37^ dollars, he should hare $555. How
much money has he ?
43. Four men bought a grindstone 4 feet in diameter, paying equal
furos, and they agreed that the first should grind off his share, then the
f econd, and so on to the last What was the thickness of the portion
ground off by each 1
44. June 1, 1852, 1 bought for cash 500 casks of oil, each cask con
taining 42 gallons, at Si. 10 per gallon. Oct 1, 1852, 1 sold it on 3
months* credit, at a price per gallon equal to 125 per cent of its cost per
gallon, deducting 5 per cent of the whole quantity of oil for leakage.
I immediately got the note received for the oil discounted at a bank.
Allowing that I paid 10 cents per cask for truckage, and $25 for storage
and other expenses, and that money was worth 6 per cent per year, did
I gain or lose, and how many dollars 1
45. A, B, and trade in company, and gain $100, of which A has
$12.50, B has $25, and C has $62.50. C put in $21 more of the origi«
nal stock tlian A and B together. What was the original stock ?
46. Keuben Aldrich and George Guild bought cloth together. Aid-
rich paying $6 more than J of its cost They sold the cloth at such
rate that Aldnch's share was $39, and Guild's share was $36. What
4id the cloth cost 1
47. Charles, John, and James were talking of their money. Charles
tas 50 dollars. James says that if Charles should give him his money,
^e should have twice as much as John ; and John says that if Charles
jhould give him his money, he should have three times as much as
James. How much has each ?
48. A speculator borrowed $2000, agreeing to pay interest at the
rate of 9 per cent per year, and invested the money in land at $80
per acre. 3 mo. afterwards he sold i the land for $900, and the rest
at $100 per acre, and expended the proceeds for flour. 2 mo. 15 da.
afterwards he sold ^ of the fiour for $600, and the remainder for what
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MISCELLANEOUS EXAM »LES. 363
he paid for the whole, and immediately paid the amount of the bor
rowed money. How much was his gain 1
49. If i of 12 were 6, what, in the same ratio, would i of 50 be ?
50. I own a square lot of land measuring 10 rods on a side. How
deep a ditch 6 feet wide must I dig around it within its limits to raise its
Burface 1 foot ?
51. A merchant sold a lot of flour at $8.40 per barrel, and thereby
gained 20 per cent. He afterwards sold another lot of the same flour
for $203, and thereby gained 16 per cent How many barrels were there
in the last lot ?
52. What must be the diameter of a sphere to contain as many cubic
inches as a cone 1 foot high and having a base of 1 foot in diameter ?
53. Mr. Hicks invested a certain sum in flour, and Mr. Gardiner in-
vested twice as much. It turned out that Mr. Hicks lost 10 per cent,
and Mr. Gardiner gained 10 per cent, and that the difference between
what Mr. Hicks received for his lot and what Mr. Gardiner received for
his was $260. How much did each invest ?
54. Jan. 1, 1852, I borrowed $954, agreeing to pay interest at the
rate of 5 per cent, and immediately expended it for cloth at $3 per yard.
Four days afterwards I sold the cloth at $3.50 per yard, to be paid Juno
17, 1852. On receipt of the money, I immediately expended it for
cloth at $1 per yard. July 1, 1852, 1 sold the cloth at $1.12^ per yard,
payable Sept. 22, 1852. As soon as this debt was paid, I put the money
on interest at 6 per cent. Jan. 1, 1853, 1 collected the amount due me,
and paid that which I owed. ' How much had I gained by the transac-
tions ?
55. Wishing to find the distance between two trees, which cannot
be directly measured on account of a swamp, I measure due east 80
yards from the foot of one of them ; then turning south, I measure
100 yards, when I find that I am just 40 yards to the east of the other
tree. How far apart are the trees 1
56. The eaves of a house are at the same height, and 30 feet apart.
The ridge pole is 12 feet higher than the eaves, and just midway be-
tween them. The house is 40 feet long. How many shingles will it
take to cover the roof, if each shingle covers a space 6 inches long
and 4 inches broad ?
57. Multiply any number by 4, add 6, divide by 2, add 7, divide by 2,
subtract 5, add twice the original number, divide by the original num-
ber, subtract 1, multiply by the original number, add 3, multiply by 3,
add 11, divide by 2, and the result will always be 10 more than 3 times
the original number. Why is this 1
68. A man spent ^ of his money and i a dollar more ; then i of what
hi had left, and ^ of a dollar more j then j- of what he had left, and i
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3C4 MISCELLANEOUS EXAMPLES.
of a dollar more, when he had jnst 7 dollars. Hew much monej hnd
ho at first 1
59. A rcctanfpilar box is 3 times as lonp: as it is wide, And twice ag
wide as it is high, and contains 96 cubic feet. How man^ square feet
are there in its surface ?
CO. Obtained at a bank, on my note payable in 3 months, money
enoufi^h to buy 20 acres of land at $100 per acre. One month after
wards I sold the land, receiving in payment a note on demand, with
ih (crest at 6 per cent per year. I collected the amount of this note tha
day my note became due at the bank, ami found that it took -]^ of it to
pay the latter. For how much per acre did I sell the land ?
61. Obtained at a bank, on my note payable in 5 months, money
ebough to buy 20 acres of land at $100 per acre, and at the same timo
hired of a friend money enough to bny another lot of the same size and
price as the first, giving in payment my note payable on demand, with
interest at 6 per cent per year. When the note at the bank became due,
I sold both lots for cash at the same price per acre, and found that the
money received for 16 acres of it was sufficient to pay my note at the
bank. How much did I gain by the transactions ? How much more on
the second lot than on the first ?
62. Bernard Far well and Francis Dana traded in company. FarwelFg
stotfk was $800, and Dana's $420. On dividing their profits, they found
that FarAVcll's share was $4 more tlian twice Dana's. How many dol-
lars did each gain ?
63. A commission merchant received on consignment 100 bags of
com from A, 1.50 bags from B, and 75 bags from C, and putting them
all into 1 lot sold them for $400. Now, allowing that A's lot is 10 per
cent better than B's. and 15 per cent better than C's, what is the just
share of each ?
64. Is the reasoning process contained in the following solution true
or false 1 If false, in what does its fallacy consist 1
Question. — "What is the eflfect of adding 3 to both numerator and
denominator of a fraction ?
Solution. — Since adding 3 to both numerator and denominator of a
fraction gives for a result a fraction which expresses 3 more parts than
the former, but of such kind that it will take 3 more of them to equal a
unit, the addition has both increased and diminished the fraction, and
has therefore not altered its value.
65. Show the fallacy, if any, in the following solution : —
If the numerator of a fraction is 4 greater than its denominator, what
will bo the effect of adding the same number to both numerator and do-
nominal or 1
Soluti m. — Adding the same number to both numerator and dcnomi'
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MISCELLANEOUS EXAMPLES. 305
ndtor of R fraction will not affect the difference between them. Hence,
th« lesuJ.ting fraction in tlie case supposed must, like the orij^inal one,
express 4 more fractional parts than it takes to equal a unit. Therefore^
the resulting fraction equals the original one, and the supposed addition
has not affected the value of the fraction.
$6. Two men, A and B, hired a horse and carriage for $7, to go from
PrcTidence to Boston and back, the distance between the cities being 42
miles. At Attleboro', 12 miles from Providence, they took in C, agree-
ing to take him to Boston and back to Attleboro' for his proportionate
share of the expense. At Walpole, 24 miles from Providence, they took
in D, agreeing to take him to Boston and back to Walpole for his T»ro-
portionate share of the expense. What ought each person to pay ?
Note. — Arithmeticians do not agree as to the correct solution of
such examples as the above ; some contending that each person should
pay in exact proportion to the number of miles he rode, and others that
A and B should each pay ^ the expense of the ride from Providence to
Attleboro', i the expense of the ride from Attleboro' to Walpole, and i
the expense of the ride from Walpole to Boston ; that C should pay -J
the expense of the ride from Attleboro' to Walpole, and ^ the expense
of the ride from Walpole to Boston ; and that D should pay only i
of the expense of the ride from Walpole to Boston. Which is the just
principle 1
67. I sold f of a lot of land for 20 per cent more than it cost, and tho
remainder for 20 per cent less than it cost. What per cent did I gain on
the whole ?
68. I sold J of a cask of wine for $36, which was 25 per cent more
than the part sold cost. I then sold the remainder at an advance of 20
per cent on its cost. What per cent of the cost of the cask did I gain 1
69. I sent to my agent in Boston a lot of flour, which he sold for
$6075, charging a commission of 2 per cent on the sales. He invested
the remainder, after deducting his commission of ij per cent on tho
purchase, in cloths, which he shipped to my agent in Savannah. Tho
latter sold them at an advance of 25 per cent on the cost, charging a
commission of 5 per cent on the sales, and invested the balance, after
deducting a commission of 2 per cent on the purchase, in cotton. Tho
cotton was shipped to my agent in Boston, who sold it at an advance of
20 per cent on it? cost, charging a commission of 1§ per cent. Allow-
ing that the expenses of freight, insurance, &c., were $1000, what was
my gain, supposing that the flour cost me $6075 ?
70. A traveller had to pass three toll gates. At the first gate he paid
5 cents less than half the money he had ; at the second he paid 2 cents
less than half of what he had left \ and at the third he paid 1 cent mor«
31*
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8M MISCELLANEOUS EXAMPLES.
Hub balf of irhat he then had, after which he fiad only 4 cents <cfk
How mach monej did he have at first 1
71. Ljman Richards and John Dexter traded in company, Bichards
pftying in $9 less than f of the whole stock. They gained $200, of
which Richard's share was $117. What was the original stock of each *
71. A farmer had his sheep in three pastures. In the first pasture there
were twice at many as in the second, and in the second twice as many
M in the third. 40 jnmped out of the first pasture into the second,
and 98 Jomped from the second into the third, when the number of sheep
in each pasture was equal. How many were originally in each pasture ?
75. A water tub holds 73 gallons. The pipe which conveys water to
it admits 7 gallons in 5 minutes, and the tap discharges 20 gallons in 17
minntes. Now, suppose that both being carelessly left open, the water
It turned on at 4 o'clock, and a servant discovers it at 6, and puts in the
tap, at what time will the tub be full ? *
74. A gentleman has two horses, and a carriage worth £100. Now,
if the first horse be harnessed in the carriage, he and the carriage
together will be worth three times as much as the second horse ; but if
tiie second horse be harnessed, he and the carriage will together be
worth 7 tUnes as much as the first What is the value of each horse I *
75. There is a fish whose head is 10 feet long ; his tail is as long as
his liead and half of his body, and his body is as long as his head and
tail together. What is the whole length of the fish ? *
76. ** If 12 oxen will eat 3^ acres of grass in 4 weeks, with all that
grows daring that time, and 21 oxen eat 10 acres in 9 weeks, with all
tliat grows during that time, how many oxen would eat 24 acres, with
the growth, in 18 weeks, the grass all the while growing uniformly ? "
77. Jan. 1, 1851, my agent in Bufialo bought for me 1000 barrels of
flour at $4 per barrel, for which he charged a commission of 1 per cent
On the dd of January, I sent him cash to pay for the flour and his com-
mission. It cost me $1 per barrel to have the flour transported to
Boston, and I incurred other expenses upon it to the amount of $20.
Feb. If 1851, 1 sold the flour to J. Smith & Co., at an advance of 25 per
cent on its entire cost, receiving in payment half cash, and their note
payable in 6 months for the remainder. I had their note discounted at
a bank ; bat before it became due they failed, so that when it became
due, I, as indorser, was obliged to pay it. Jan. 1, 1852, 1 settled with J.
Smith & Co., receiving 50 cents on each dollar they owed me. Allow-
ing that money was worth 6 per cent per year, and that I paid the
firaight and other expenses of the flour on the 15th of January, what
was the amount of my loss 1
■ /^yyci-
* From Pike's Arithmetic, edition of 1788.
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